Maths-
General
Easy

Question

In how many ways can six different rings be wear in four fingers?

  1. 6P4    
  2. 64    
  3. 46    
  4. 6C4

Hint:

We will first start by finding the way in which one ring can be worn in 4 fingers. Then we will do the same for 6 rings and then using the fundamental principle of counting we will find the total ways.

The correct answer is: 46


    Detailed Solution
    Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
    Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
    Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.
    Now, we know that by the fundamental principle of counting there can be 4 cross times 4 cross times 4 cross times 4 cross times 4 cross times 4 ways of wearing 6 rings.
    S o comma space w e space h a v e space 4 to the power of 6 equals 4096 space w a y s space t o space w e a r space 6 space d i f f e r e n t space t y p e s space o f space r i n g s.

    It is important to note that we have used a basic fundamental principle of counting to find the total ways. Also, it is important to notice that each ring has 4 ways as it has not been given that each finger must have at least one ring. So, there can be 6 rings in a finger alone and remaining all the fingers empty.

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    How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

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    The number of ways in which three persons can dress themselves when they have 4 shirts. 5 pants and 6 hats between them, is-

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    Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-

    Eleven animals of a circus have to be placed in eleven cages, one in each cage. If four of the cages are too small for six of the animals, the number of ways of caging the animals is-

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    Eight chairs are numbered from 1 to 8. Two women and three men wish to occupy one chair each. First women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from the remaining. The number of possible arrangements is-

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    A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

     DETAILED SOLUTION:

    There are 16 people for the tea party.
    People sit along a long table with 8 chairs on each side.
    Out of 16, 4 people sit on a particular side and 2 sit on the other side.
    Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.

    Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
    The number of ways of choosing 6 people out of 10 are C presuperscript 10 subscript 6
    Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are C presuperscript 4 subscript 4 space end subscript equals space 1
    And now all the 16 people are placed in their seats according to the constraints.

    Now we have to arrange them.
    So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is(8!×8!).

    So, a possible number of arrangements will be

    rightwards double arrow C presuperscript 10 subscript 6 space end subscript cross times space C presuperscript 4 subscript 4 space end subscript cross times space 8 factorial thin space cross times 8 factorial

    Now as we know


    C presuperscript n subscript r space end subscript equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction
S o comma space C presuperscript 10 subscript 6 space end subscript equals space fraction numerator 10 factorial over denominator 4 factorial space cross times 6 factorial end fraction equals space fraction numerator 10.9.8.7.6 factorial over denominator 4.3.2.1.6 factorial end fraction equals space 210
A n d space C presuperscript 4 subscript 4 space end subscript equals space fraction numerator 4 factorial over denominator 4 factorial thin space cross times 0 factorial end fraction equals space 1

    So total number of arrangements is
    210×(8× 8!)

    A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

    Maths-General
     DETAILED SOLUTION:

    There are 16 people for the tea party.
    People sit along a long table with 8 chairs on each side.
    Out of 16, 4 people sit on a particular side and 2 sit on the other side.
    Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.

    Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
    The number of ways of choosing 6 people out of 10 are C presuperscript 10 subscript 6
    Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are C presuperscript 4 subscript 4 space end subscript equals space 1
    And now all the 16 people are placed in their seats according to the constraints.

    Now we have to arrange them.
    So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is(8!×8!).

    So, a possible number of arrangements will be

    rightwards double arrow C presuperscript 10 subscript 6 space end subscript cross times space C presuperscript 4 subscript 4 space end subscript cross times space 8 factorial thin space cross times 8 factorial

    Now as we know


    C presuperscript n subscript r space end subscript equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction
S o comma space C presuperscript 10 subscript 6 space end subscript equals space fraction numerator 10 factorial over denominator 4 factorial space cross times 6 factorial end fraction equals space fraction numerator 10.9.8.7.6 factorial over denominator 4.3.2.1.6 factorial end fraction equals space 210
A n d space C presuperscript 4 subscript 4 space end subscript equals space fraction numerator 4 factorial over denominator 4 factorial thin space cross times 0 factorial end fraction equals space 1

    So total number of arrangements is
    210×(8× 8!)
    General
    Maths-

    If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

    If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

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    A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

    W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript
    equals negative U subscript P end subscript equals negative m V subscript P end subscript
    equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses
    Potential at point P will be obtained by in integration as given below. Let d M be the mass of small rings as shown

    d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r
    equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction
    d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction
    equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r
    equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses
    therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

    A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

    physics-General
    W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript
    equals negative U subscript P end subscript equals negative m V subscript P end subscript
    equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses
    Potential at point P will be obtained by in integration as given below. Let d M be the mass of small rings as shown

    d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r
    equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction
    d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction
    equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r
    equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses
    therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis
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    The two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

    In the pulley arrangement open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g
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    stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction
    equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g

    The two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

    physics-General
    In the pulley arrangement open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g
    But stack a with rightwards arrow on top subscript 1 end subscript is in downward direction and in the upward direction i e, stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript
    therefore Acceleration of centre of mass
    stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction
    equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g
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    If x equals 1 plus 3 a plus 6 a squared plus 10 a cubed plus midline horizontal ellipsis. to straight infinity terms, vertical line a vertical line less than 1 comma y equals 1 plus 4 a plus 10 a squared plus 20 a cubed plus midline horizontal ellipsis straight infinity terms, vertical line a vertical line less than 1, then x colon y

    If x equals 1 plus 3 a plus 6 a squared plus 10 a cubed plus midline horizontal ellipsis. to straight infinity terms, vertical line a vertical line less than 1 comma y equals 1 plus 4 a plus 10 a squared plus 20 a cubed plus midline horizontal ellipsis straight infinity terms, vertical line a vertical line less than 1, then x colon y

    maths-General
    parallel
    General
    Maths-

    The coefficient of x to the power of negative n end exponent in left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent is

    The coefficient of x to the power of negative n end exponent in left parenthesis 1 plus x right parenthesis to the power of n end exponent open parentheses 1 plus fraction numerator 1 over denominator x end fraction close parentheses to the power of n end exponent is

    Maths-General
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    maths-

    open parentheses 1 plus x plus x squared plus horizontal ellipsis plus x to the power of p close parentheses to the power of n equals a subscript 0 plus a subscript 1 x plus a subscript 2 x squared plus horizontal ellipsis plus a subscript n p end subscript x to the power of n p end exponent not stretchy rightwards double arrow a subscript 1 plus 2 a subscript 2 plus 3 a subscript 3 plus horizontal ellipsis plus n p

    open parentheses 1 plus x plus x squared plus horizontal ellipsis plus x to the power of p close parentheses to the power of n equals a subscript 0 plus a subscript 1 x plus a subscript 2 x squared plus horizontal ellipsis plus a subscript n p end subscript x to the power of n p end exponent not stretchy rightwards double arrow a subscript 1 plus 2 a subscript 2 plus 3 a subscript 3 plus horizontal ellipsis plus n p

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    General
    chemistry-

    Compounds (A) and (B) are – 

    Compounds (A) and (B) are – 

    chemistry-General
    parallel
    General
    Maths-

    2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals

    2. C subscript 0 plus 5. C subscript 1 plus 8 C subscript 2 plus........... plus left parenthesis 3 n plus 4 right parenthesis C subscript n
equals sum from r equals 0 to n of left parenthesis 3 n plus 4 right parenthesis space C subscript r
equals sum from r equals 0 to n of 3 n. space C subscript r plus sum from r equals 0 to n of 2. C subscript r space space space space space space space space space space space space space space space open square brackets a s space r equals 0 comma space s o comma space space C subscript r equals 0 close square brackets
equals sum from r equals 1 to n of 3 n. space C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript plus sum from r equals 0 to n of 2. C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript space space space space space space space space space space space open square brackets a s space 2 to the power of n equals C presuperscript n subscript 0 plus C presuperscript n subscript 1 plus C presuperscript n subscript 2 plus C presuperscript n subscript 3......... plus C presuperscript n subscript n close square brackets
equals 3 n.2 to the power of n minus 1 end exponent plus 2.2 to the power of n minus 1 end exponent
equals left parenthesis 2 plus 3 n right parenthesis 2 to the power of n minus 1 end exponent space space space space

    2 times C subscript 0 plus 5 times C subscript 1 plus 8 times C subscript 2 plus horizontal ellipsis plus left parenthesis 2 plus 3 n right parenthesis times C subscript n equals

    Maths-General
    2. C subscript 0 plus 5. C subscript 1 plus 8 C subscript 2 plus........... plus left parenthesis 3 n plus 4 right parenthesis C subscript n
equals sum from r equals 0 to n of left parenthesis 3 n plus 4 right parenthesis space C subscript r
equals sum from r equals 0 to n of 3 n. space C subscript r plus sum from r equals 0 to n of 2. C subscript r space space space space space space space space space space space space space space space open square brackets a s space r equals 0 comma space s o comma space space C subscript r equals 0 close square brackets
equals sum from r equals 1 to n of 3 n. space C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript plus sum from r equals 0 to n of 2. C presuperscript n minus 1 end presuperscript subscript r minus 1 end subscript space space space space space space space space space space space open square brackets a s space 2 to the power of n equals C presuperscript n subscript 0 plus C presuperscript n subscript 1 plus C presuperscript n subscript 2 plus C presuperscript n subscript 3......... plus C presuperscript n subscript n close square brackets
equals 3 n.2 to the power of n minus 1 end exponent plus 2.2 to the power of n minus 1 end exponent
equals left parenthesis 2 plus 3 n right parenthesis 2 to the power of n minus 1 end exponent space space space space
    General
    maths-

    A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

    A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to:

    maths-General
    General
    Maths-

    If one root of the equation a x squared plus b x plus c equals 0 is reciprocal of the one of the roots of equation  a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 then

    Let one of the roots of the equation a x squared plus b x plus c equals 0 be alpha.
    Then as per question one of the roots of the equation a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 will be1 over alpha.
    a x squared plus b x plus c equals 0
rightwards double arrow a alpha squared plus b alpha plus c equals 0 space _ _ _ _ _ _ _ _ _ e q u a t i o n space 1
    a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0
rightwards double arrow a subscript 1 open parentheses 1 over alpha close parentheses squared plus b subscript 1 open parentheses 1 over alpha close parentheses plus c subscript 1 equals 0 _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 2
    Multiplying equation 1 by c subscript 1 and equation 2 by a, then subtracting, we get:
    stack attributes charalign center stackalign right end attributes row a c subscript 1 alpha squared plus b c subscript 1 alpha plus c c subscript 1 equals 0 end row row a c subscript 1 alpha squared plus a b subscript 1 alpha plus a subscript 1 a equals 0 end row row minus none end row horizontal line row b c subscript 1 alpha minus a b subscript 1 alpha plus c c subscript 1 minus a subscript 1 a equals 0 end row end stack
    rightwards double arrow alpha open parentheses b c subscript 1 minus b subscript 1 a close parentheses plus c c subscript 1 minus a subscript 1 a equals 0
rightwards double arrow alpha equals fraction numerator a subscript 1 a minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction
    putting the value in equation 1
    a open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses squared plus b open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses plus c equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus b open parentheses a a subscript 1 minus c c subscript 1 close parentheses open parentheses b c subscript 1 minus a b subscript 1 close parentheses plus c open parentheses b c subscript 1 minus a b subscript 1 close parentheses squared equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses open parentheses a a subscript 1 b minus c c subscript 1 b plus a b subscript 1 c plus b c c subscript 1 close parentheses equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals negative open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis
rightwards double arrow open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus a b subscript 1 close parentheses left parenthesis b subscript 1 c minus a subscript 1 b right parenthesis

    If one root of the equation a x squared plus b x plus c equals 0 is reciprocal of the one of the roots of equation  a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 then

    Maths-General
    Let one of the roots of the equation a x squared plus b x plus c equals 0 be alpha.
    Then as per question one of the roots of the equation a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0 will be1 over alpha.
    a x squared plus b x plus c equals 0
rightwards double arrow a alpha squared plus b alpha plus c equals 0 space _ _ _ _ _ _ _ _ _ e q u a t i o n space 1
    a subscript 1 x squared plus b subscript 1 x plus c subscript 1 equals 0
rightwards double arrow a subscript 1 open parentheses 1 over alpha close parentheses squared plus b subscript 1 open parentheses 1 over alpha close parentheses plus c subscript 1 equals 0 _ _ _ _ _ _ _ _ _ _ _ e q u a t i o n space 2
    Multiplying equation 1 by c subscript 1 and equation 2 by a, then subtracting, we get:
    stack attributes charalign center stackalign right end attributes row a c subscript 1 alpha squared plus b c subscript 1 alpha plus c c subscript 1 equals 0 end row row a c subscript 1 alpha squared plus a b subscript 1 alpha plus a subscript 1 a equals 0 end row row minus none end row horizontal line row b c subscript 1 alpha minus a b subscript 1 alpha plus c c subscript 1 minus a subscript 1 a equals 0 end row end stack
    rightwards double arrow alpha open parentheses b c subscript 1 minus b subscript 1 a close parentheses plus c c subscript 1 minus a subscript 1 a equals 0
rightwards double arrow alpha equals fraction numerator a subscript 1 a minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction
    putting the value in equation 1
    a open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses squared plus b open parentheses fraction numerator a a subscript 1 minus c c subscript 1 over denominator b c subscript 1 minus a b subscript 1 end fraction close parentheses plus c equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus b open parentheses a a subscript 1 minus c c subscript 1 close parentheses open parentheses b c subscript 1 minus a b subscript 1 close parentheses plus c open parentheses b c subscript 1 minus a b subscript 1 close parentheses squared equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses open parentheses a a subscript 1 b minus c c subscript 1 b plus a b subscript 1 c plus b c c subscript 1 close parentheses equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared plus open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis equals 0
rightwards double arrow a open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals negative open parentheses b c subscript 1 minus a b subscript 1 close parentheses a left parenthesis a subscript 1 b minus b subscript 1 c right parenthesis
rightwards double arrow open parentheses a a subscript 1 minus c c subscript 1 close parentheses squared equals open parentheses b c subscript 1 minus a b subscript 1 close parentheses left parenthesis b subscript 1 c minus a subscript 1 b right parenthesis
    parallel

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