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Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -

Maths-General

  1. 1800    
  2. 2400    
  3. 1200    
  4. 3000    

    Answer:The correct answer is: 2400For 1 less or equal thanless or equal than 4, let xi (greater or equal than 3) be the number of blanks between ith and (i + 1)th letters. Then,
    x1 + x2 + x3 + x4 = 15 …..(1)
    The number of solutions of (1)
    = coefficient of t15 in (t3 + t4 +….)4
    = coefficient of t3 in (1 – t)–4
    = coefficient of t3 in [1 + 4C1 + 5C2 t2 + 6C3 t3 + …..]
    = 6C3 = 20.
    But 5 letters can be permuted in 5! = 120 ways.
    Thus, the required number arrangements
    = (120) (20) = 2400.
    Hence

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    Related Questions to study

    General
    maths-

    The number of ordered pairs (m, n), m, n element of {1, 2, … 100} such that 7m + 7n is divisible by 5 is -

    Note that 7r (r  N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
    Thus, 7m + 7n cannot end in 5 for any values of m, n  N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
    For 7m + 7n to end in 0, the forms of m and n should be as follows :
    m n
    1 4r 4s + 2
    2 4r + 1 4s + 3
    3 4r + 2 4s
    4 4r + 3 4s + 1
    Thus, for a given value of m there are just 25 values of n for which 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
     There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5.
    Hence

    The number of ordered pairs (m, n), m, n element of {1, 2, … 100} such that 7m + 7n is divisible by 5 is -

    maths-General
    Note that 7r (r  N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
    Thus, 7m + 7n cannot end in 5 for any values of m, n  N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
    For 7m + 7n to end in 0, the forms of m and n should be as follows :
    m n
    1 4r 4s + 2
    2 4r + 1 4s + 3
    3 4r + 2 4s
    4 4r + 3 4s + 1
    Thus, for a given value of m there are just 25 values of n for which 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
     There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5.
    Hence
    General
    maths-

    Consider the following statements:
    1. The number of ways of arranging m different things taken all at a time in which p less or equal than m particular things are never together is m! – (m – p + 1)! p!.
    2. A pack of 52 cards can be divided equally among four players in order in fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fractionways.
    Which of these is/are correct?

    (1) Total number of ways of arranging m things = m!.
    To find the number of ways in which p particular things are together, we consider p particular things as a group.
     Number of ways in which p particular things are together = (m – p + 1)! p!
    So, number of ways in which p particular things are not together
    = m! – (m – p + 1)! p!
    Total number of ways = fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction
    Hence, both of statements are correct.

    Consider the following statements:
    1. The number of ways of arranging m different things taken all at a time in which p less or equal than m particular things are never together is m! – (m – p + 1)! p!.
    2. A pack of 52 cards can be divided equally among four players in order in fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fractionways.
    Which of these is/are correct?

    maths-General
    (1) Total number of ways of arranging m things = m!.
    To find the number of ways in which p particular things are together, we consider p particular things as a group.
     Number of ways in which p particular things are together = (m – p + 1)! p!
    So, number of ways in which p particular things are not together
    = m! – (m – p + 1)! p!
    Total number of ways = fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction
    Hence, both of statements are correct.
    General
    maths-

    The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) less or equal than ƒ(j), straight for all i < j, is equal to-

    Let ‘l’ is associated with ‘r’ ,
    element of {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
    Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
    = not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript open parentheses not stretchy sum subscript j equals r end subscript superscript 5 end superscript left parenthesis 6 minus j right parenthesis close parentheses= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript fraction numerator left parenthesis 6 minus r right parenthesis left parenthesis 7 minus r right parenthesis over denominator 2 end fraction
    = fraction numerator 1 over denominator 2 end fraction open parentheses not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript left parenthesis 42 minus 13 r plus r to the power of 2 end exponent right parenthesis close parentheses
    = fraction numerator 1 over denominator 2 end fraction open parentheses 42.5 minus 13. fraction numerator 6.5 over denominator 2 end fraction plus fraction numerator 5.6.11 over denominator 6 end fraction close parentheses= 35
    Hence (a) is correct answer.

    The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) less or equal than ƒ(j), straight for all i < j, is equal to-

    maths-General
    Let ‘l’ is associated with ‘r’ ,
    element of {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.
    Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions
    = not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript open parentheses not stretchy sum subscript j equals r end subscript superscript 5 end superscript left parenthesis 6 minus j right parenthesis close parentheses= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript fraction numerator left parenthesis 6 minus r right parenthesis left parenthesis 7 minus r right parenthesis over denominator 2 end fraction
    = fraction numerator 1 over denominator 2 end fraction open parentheses not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript left parenthesis 42 minus 13 r plus r to the power of 2 end exponent right parenthesis close parentheses
    = fraction numerator 1 over denominator 2 end fraction open parentheses 42.5 minus 13. fraction numerator 6.5 over denominator 2 end fraction plus fraction numerator 5.6.11 over denominator 6 end fraction close parentheses= 35
    Hence (a) is correct answer.
    General
    maths-

    The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x|  less or equal than k, |y| less or equal than k, |x – y| less or equal than k ; is-

    |x| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(1)
    & |y| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(2)
    & |x – y| less or equal thannot stretchy rightwards double arrow |y – x| less or equal than k ….(3)

    not stretchy rightwards double arrow – k less or equal than y – x less or equal than k not stretchy rightwards double arrow x – k less or equal than y less or equal than x + k
    thereforeNumber of points having integral coordinates
    = (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
    = (3k2 + 3k + 1).

    The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x|  less or equal than k, |y| less or equal than k, |x – y| less or equal than k ; is-

    maths-General
    |x| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(1)
    & |y| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(2)
    & |x – y| less or equal thannot stretchy rightwards double arrow |y – x| less or equal than k ….(3)

    not stretchy rightwards double arrow – k less or equal than y – x less or equal than k not stretchy rightwards double arrow x – k less or equal than y less or equal than x + k
    thereforeNumber of points having integral coordinates
    = (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
    = (3k2 + 3k + 1).
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    maths-

    The numbers of integers between 1 and 106 have the sum of their digit equal to K(where 0 < K < 18) is -

    The required no. of ways = no. of solution of the equation (x1 + x2 + x3 + x4 + x5 + x6 = K)
    Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
    = Coefficient of xK in (1 + x + x2 +….. + x9)6
    = Coefficient of xK in open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
    = Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
    (1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
    = k + 6CK – 6. K–4CK–10
    = k + 6C6 – 6. K–4C6 .

    The numbers of integers between 1 and 106 have the sum of their digit equal to K(where 0 < K < 18) is -

    maths-General
    The required no. of ways = no. of solution of the equation (x1 + x2 + x3 + x4 + x5 + x6 = K)
    Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
    = Coefficient of xK in (1 + x + x2 +….. + x9)6
    = Coefficient of xK in open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
    = Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
    (1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
    = k + 6CK – 6. K–4CK–10
    = k + 6C6 – 6. K–4C6 .
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    maths-

    The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. The maximum number of triangles formed with vertices at these points are -

    Total number of points = m +n + k. Therefore the total number of triangles formed by these points is m + n + kC3. But out of these m + n + k points, m points lie on I1, n points lie on I2 and k points lie on I3 and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
    m + n + kC3mC3nC3kC3.

    The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. The maximum number of triangles formed with vertices at these points are -

    maths-General
    Total number of points = m +n + k. Therefore the total number of triangles formed by these points is m + n + kC3. But out of these m + n + k points, m points lie on I1, n points lie on I2 and k points lie on I3 and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
    m + n + kC3mC3nC3kC3.
    General
    maths-

    If the line l x plus m y equals 1 is a normal to the hyperbola fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 then fraction numerator a to the power of 2 end exponent over denominator l to the power of 2 end exponent end fraction minus fraction numerator b to the power of 2 end exponent over denominator m to the power of 2 end exponent end fraction equals

    If the line l x plus m y equals 1 is a normal to the hyperbola fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 then fraction numerator a to the power of 2 end exponent over denominator l to the power of 2 end exponent end fraction minus fraction numerator b to the power of 2 end exponent over denominator m to the power of 2 end exponent end fraction equals

    maths-General
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    maths-

    If the tangents drawn from a point on the hyperbola x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 make angles α and β with the transverse axis of the hyperbola, then

    If the tangents drawn from a point on the hyperbola x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 make angles α and β with the transverse axis of the hyperbola, then

    maths-General
    General
    maths-

    If a hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 as its asymptotes, then the equation of its tranvsverse and conjugate axes are

    If a hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 as its asymptotes, then the equation of its tranvsverse and conjugate axes are

    maths-General
    General
    physics-

    Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line O A in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line O B. If the frequency of P is 341 H z comma the frequency of Q will be

    n subscript Q end subscript equals 341 plus-or-minus 3 equals 344 H z or 338 H z
    On waxing Q comma the number of beats decreases hence
    n subscript Q end subscript equals 344 H z

    Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line O A in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line O B. If the frequency of P is 341 H z comma the frequency of Q will be

    physics-General
    n subscript Q end subscript equals 341 plus-or-minus 3 equals 344 H z or 338 H z
    On waxing Q comma the number of beats decreases hence
    n subscript Q end subscript equals 344 H z
    General
    maths-

    The cartesian equation of r equals a s i n space 2 theta is

    The cartesian equation of r equals a s i n space 2 theta is

    maths-General
    General
    maths-

    The polar equation of y to the power of 2 end exponent equals 4 x is

    The polar equation of y to the power of 2 end exponent equals 4 x is

    maths-General
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    maths-

    The equation of the line passing through pole and open parentheses 2 comma fraction numerator pi over denominator 3 end fraction close parentheses is

    The equation of the line passing through pole and open parentheses 2 comma fraction numerator pi over denominator 3 end fraction close parentheses is

    maths-General
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    maths-

    The polar equation of the straight line passing through open parentheses 2 comma fraction numerator pi over denominator 6 end fraction close parentheses and parallel to the initial line is

    The polar equation of the straight line passing through open parentheses 2 comma fraction numerator pi over denominator 6 end fraction close parentheses and parallel to the initial line is

    maths-General
    General
    maths-

    The polar equation of the straight line passing through open parentheses 6 comma fraction numerator pi over denominator 3 end fraction close parentheses and perpendicular to the initial line is

    The polar equation of the straight line passing through open parentheses 6 comma fraction numerator pi over denominator 3 end fraction close parentheses and perpendicular to the initial line is

    maths-General