General
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Maths-

A double decker bus can accommodate (u + ell) passengers, u in the upper deck and ell in the lower deck. The number of ways in which the (u + ell) passengers can be distributed in the two decks, if r (less or equal than ell) particular passengers refuse to go in the upper deck and S(less or equal thanu) refuse to sit in the lower deck, is -

Maths-General

  1. fraction numerator left parenthesis u plus l – r – S right parenthesis factorial over denominator left parenthesis l – r right parenthesis factorial left parenthesis u – S right parenthesis factorial end fraction    
  2. u–SPell–r    
  3. fraction numerator left parenthesis u plus l right parenthesis factorial over denominator r factorial S factorial end fraction    
  4. u–SCell–r    

    Answer:The correct answer is: fraction numerator left parenthesis u plus l – r – S right parenthesis factorial over denominator left parenthesis l – r right parenthesis factorial left parenthesis u – S right parenthesis factorial end fraction

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    Related Questions to study

    General
    maths-

    The largest value of r satisfying inequality 20Cr greater or equal than 320Cr – 1 is -

    The largest value of r satisfying inequality 20Cr greater or equal than 320Cr – 1 is -

    maths-General
    General
    maths-

    Domain of f(x) = log10(log10(1 + x3)) is -

    because log10
    (1 + x3) > 0 Take antilog
    1+ x3 > 1 not stretchy rightwards double arrow x3 > 0 therefore x > 0
    element of (0, straight infinity) or x element of R+

    Domain of f(x) = log10(log10(1 + x3)) is -

    maths-General
    because log10
    (1 + x3) > 0 Take antilog
    1+ x3 > 1 not stretchy rightwards double arrow x3 > 0 therefore x > 0
    element of (0, straight infinity) or x element of R+
    General
    maths-

    If f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus vertical line x vertical line end fraction for x element of R, then f to the power of straight prime left parenthesis 0 right parenthesis equals

    If f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus vertical line x vertical line end fraction for x element of R, then f to the power of straight prime left parenthesis 0 right parenthesis equals

    maths-General
    General
    maths-

    The maximum value of P such that 3P divides 99 × 97 × 95 × .... × 51 is-

    99 × 97 × 95 ×....× 51
    equals times fraction numerator 100 factorial over denominator left parenthesis 100 cross times 98 cross times 96 cross times horizontal ellipsis cross times 52 right parenthesis end fraction cross times fraction numerator 1 over denominator 50 factorial end fraction times fraction numerator 100 factorial over denominator 2 to the power of 25 cross times 50 factorial cross times 50 factorial end fraction
    maximum power of 3 in 100 !
    equals times open square brackets 100 over 3 close square brackets plus open square brackets 100 over 9 close square brackets plus open square brackets 100 over 27 close square brackets plus open square brackets 100 over 81 close square brackets
    = 33 + 11 + 3 + 1 = 48.
    maximum power of 3 in 50! = open ceil 50 over 3 close ceil plus open ceil 50 over 9 close ceil plus open ceil 50 over 27 close ceil
    = 16 + 5 + 1 = 22
    maximum power of 3 in 25! = open square brackets 25 over 3 close square brackets plus open ceil 25 over 9 close square brackets
    = 8 + 2 = 10
    thereforeexponent of 3 = 48 + 10 –
    (22 × 2) = 14

    The maximum value of P such that 3P divides 99 × 97 × 95 × .... × 51 is-

    maths-General
    99 × 97 × 95 ×....× 51
    equals times fraction numerator 100 factorial over denominator left parenthesis 100 cross times 98 cross times 96 cross times horizontal ellipsis cross times 52 right parenthesis end fraction cross times fraction numerator 1 over denominator 50 factorial end fraction times fraction numerator 100 factorial over denominator 2 to the power of 25 cross times 50 factorial cross times 50 factorial end fraction
    maximum power of 3 in 100 !
    equals times open square brackets 100 over 3 close square brackets plus open square brackets 100 over 9 close square brackets plus open square brackets 100 over 27 close square brackets plus open square brackets 100 over 81 close square brackets
    = 33 + 11 + 3 + 1 = 48.
    maximum power of 3 in 50! = open ceil 50 over 3 close ceil plus open ceil 50 over 9 close ceil plus open ceil 50 over 27 close ceil
    = 16 + 5 + 1 = 22
    maximum power of 3 in 25! = open square brackets 25 over 3 close square brackets plus open ceil 25 over 9 close square brackets
    = 8 + 2 = 10
    thereforeexponent of 3 = 48 + 10 –
    (22 × 2) = 14
    General
    maths-

    The number of five-digit telephone numbers having at least one of their digits repeated is

    The number of five-digit telephone numbers which can be formed using the digits 0, 1, 2 ….9 is 105 . The number of five-digit telephone number which have none of their digits repeated is 10P5 = 30240. Thus, the required number of telephone number is 105 – 30240 = 69760.

    The number of five-digit telephone numbers having at least one of their digits repeated is

    maths-General
    The number of five-digit telephone numbers which can be formed using the digits 0, 1, 2 ….9 is 105 . The number of five-digit telephone number which have none of their digits repeated is 10P5 = 30240. Thus, the required number of telephone number is 105 – 30240 = 69760.
    General
    maths-

    One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways  of doing this is -

    Let the children be A,B.C
    A   B   C  open table attributes columnalign right end attributes row 202159 row 3941 end table close curly brackets→19
    text  .................  end text
    323335 right curly bracket not stretchy rightwards arrow 1
    Total number of ways
    equals 1 plus 2 plus midline horizontal ellipsis.. plus 19 minus left parenthesis 17 plus 14 plus 11 plus 8 plus 5 plus 2 right parenthesis
    equals 190 minus 57 equals 133
    By termination
    Permutation
    133 cross times 3 factorial equals 798

    One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways  of doing this is -

    maths-General
    Let the children be A,B.C
    A   B   C  open table attributes columnalign right end attributes row 202159 row 3941 end table close curly brackets→19
    text  .................  end text
    323335 right curly bracket not stretchy rightwards arrow 1
    Total number of ways
    equals 1 plus 2 plus midline horizontal ellipsis.. plus 19 minus left parenthesis 17 plus 14 plus 11 plus 8 plus 5 plus 2 right parenthesis
    equals 190 minus 57 equals 133
    By termination
    Permutation
    133 cross times 3 factorial equals 798
    General
    maths-

    Number of ways in which 25 identical balls can be distributed among Ram, Shyam, Sunder and Ghanshyam such that atleast 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and Ghanshyam respectively, is-

    First of all select 1 + 2 + 3 + 4 = 10 balls out of 25 identical balls and distribute them as desired. It can happen only in one way. Now let x1, x2, x3 and x4 balls are given to them respectively.
    (where x1, x2, x3, x4 greater or equal than 0)
    Its now one can get any number of balls
    therefore non negative integral solution of
    x1 + x2 + x3 + x4 = 15
    will be the number of ways so
    15 + 4 –1C4–1 = 18C3

    Number of ways in which 25 identical balls can be distributed among Ram, Shyam, Sunder and Ghanshyam such that atleast 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and Ghanshyam respectively, is-

    maths-General
    First of all select 1 + 2 + 3 + 4 = 10 balls out of 25 identical balls and distribute them as desired. It can happen only in one way. Now let x1, x2, x3 and x4 balls are given to them respectively.
    (where x1, x2, x3, x4 greater or equal than 0)
    Its now one can get any number of balls
    therefore non negative integral solution of
    x1 + x2 + x3 + x4 = 15
    will be the number of ways so
    15 + 4 –1C4–1 = 18C3
    General
    maths-

    A seven digit number is in form of abcdefg (g, f, e, etc. are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are

    Cases : i) If d = 6 then seven digit numbers possible are = 5C3. 3C3
    [as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
    ii) If d = 7 then numbers possible =,6 straight C subscript 3 times to the power of 4 straight C subscript 3
    iii) If d = 8 then numbers possible = 7 C subscript 3 times to the power of 5 C subscript 3
    iv) If d = 9 then numbers possible = 8 straight C subscript 3 times to the power of 6 straight C subscript 3
    Add all cases

    A seven digit number is in form of abcdefg (g, f, e, etc. are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are

    maths-General
    Cases : i) If d = 6 then seven digit numbers possible are = 5C3. 3C3
    [as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
    ii) If d = 7 then numbers possible =,6 straight C subscript 3 times to the power of 4 straight C subscript 3
    iii) If d = 8 then numbers possible = 7 C subscript 3 times to the power of 5 C subscript 3
    iv) If d = 9 then numbers possible = 8 straight C subscript 3 times to the power of 6 straight C subscript 3
    Add all cases
    General
    chemistry-

    The correct orders about compounds I and II are:
    i) 
    ii) 

    The correct orders about compounds I and II are:
    i) 
    ii) 

    chemistry-General
    General
    chemistry-

    Which of the following is correct method for separating a mixture of following compounds?

    Which of the following is correct method for separating a mixture of following compounds?

    chemistry-General
    General
    chemistry-

    The order in which the reagent must be used to separate the compound I - IV is:

    The order in which the reagent must be used to separate the compound I - IV is:

    chemistry-General
    General
    maths-

    Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy

    therefore5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.
    So this can be done is
    equals fraction numerator 5 straight C subscript 2 times cubed straight C subscript 1 cross times squared straight C subscript 1 cross times to the power of 1 straight C subscript 1 cross times 4 factorial over denominator 3 factorial end fraction times open square brackets table attributes columnalign left columnspacing 1em end attributes row cell text  asgroup  end text end cell row cell text  willbe  end text 2 comma 1 comma 1 comma 1 end cell end table close square brackets
    = 10 × 3 × 2 × 4 = 240
    Now one child can be rejected is 5 straight C subscript 1 = 5 ways
    thereforeTotal ways = 5 × 240 = 1200

    Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy

    maths-General
    therefore5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.
    So this can be done is
    equals fraction numerator 5 straight C subscript 2 times cubed straight C subscript 1 cross times squared straight C subscript 1 cross times to the power of 1 straight C subscript 1 cross times 4 factorial over denominator 3 factorial end fraction times open square brackets table attributes columnalign left columnspacing 1em end attributes row cell text  asgroup  end text end cell row cell text  willbe  end text 2 comma 1 comma 1 comma 1 end cell end table close square brackets
    = 10 × 3 × 2 × 4 = 240
    Now one child can be rejected is 5 straight C subscript 1 = 5 ways
    thereforeTotal ways = 5 × 240 = 1200
    General
    chemistry-

    Decreasing order of solubility of following compounds is:
    i) 
    ii) 
    iii) 
    iv) 

    Decreasing order of solubility of following compounds is:
    i) 
    ii) 
    iii) 
    iv) 

    chemistry-General
    General
    chemistry-

    Which of the following statement is correct about tropolone?

    Which of the following statement is correct about tropolone?

    chemistry-General
    General
    chemistry-

    In which case first has higher solubility than second?
    I) Phenol, Benzene
    II) Nitrobenzene, Phenol
    III) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehyde
    IV) CH3CHO, CH3–O–CH3
    V) o–Nitrophenol, p–Nitrophenol
    VI) 

    In which case first has higher solubility than second?
    I) Phenol, Benzene
    II) Nitrobenzene, Phenol
    III) o–Hydroxybenzaldehyde, p–Hydroxy benzaldehyde
    IV) CH3CHO, CH3–O–CH3
    V) o–Nitrophenol, p–Nitrophenol
    VI) 

    chemistry-General