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The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives atleast 4 and atmost 7 candies, C3 receives atleast 2 and atmost 6 candies, is equal to:

Option 1 -

205

Option 2 -

615

Option 3 -

510

Option 4 -

430

0 5 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 4


    Detailed Solution:

    Let the number of chocolates given to C1, C2, C3 & C4 be a, b, c, d respectively.

    Given 4   ≤ b ≤ 7

    2 ≤ c ≤ 6

    Now using these the maximum number of chocolates that can be given to C1 or C4 is 24 (where b & c are given 2 & 4 chocolates).

    ∴ 0 ≤ a ≤ 2 4

    0 ≤ d ≤ 2 4

    & a + b + c + d = 30

    So, total possible solution to the above equation.

    Coefficient of x30 in.

    ( x 0 + x + x 2 + . . . . + x 2 4 ) ( x 4 + x 5 + . . . . + x 7 ) ( x 2 + x 3 + . . . . + x 6 ) ( x 0 + x + x 2 + . . . . + x 2 4 )

    =  (1+....+x24)2(x4)(1+x+....+x3)×x2(1+x+....+x4)

    = ( x 5 6 − 2 x 3 1 + x 6 ) × ( x 9 − x 4 − x 5 + 1 ) × ( x − 1 ) − 4

    x56 & x31 can never give x30 so we discard them.

    x 6 × x 9 × ( x − 1 ) − 4 → 1 5 + 4 − 1 ? C 4 − 1 = 1 8 ? C 3

    Coefficient x30 ® 18C3 – 23C3 – 22C3 + 27C3

    =   1 8 × 1 7 × 1 6 6 − 2 3 × 2 2 × 2 1 6 − 2 2 × 2 1 × 2 0 6 + 2 7 × 2 6 × 2 5 6

    = 430

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