Distributing

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The number of ways of distributing six different books amongst three persons such that a particular person gets three or more books is:


The answer is 233.

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1 solution

Vishnu Kadiri
Jun 7, 2020

If person 1 is to get books 1, 2, 3, then he will receive it only 1 way and not 3 ! 3! ways. There are two methods to find the answer.

Method 1: Arrange the six books in a line. This can be done in 6 ! 6! ways. Now, choose one of the arrangements, say 245316. Now this can be distributed in the following way: 24|5|16, where first partition goes to person 1, second to second, and third to third. So, if 1 receives 2 books and 2 receives 1, 3 receives 3 books, this can be done in 6 ! 2 ! 1 ! 3 ! \frac{6!}{2!1!3!} ways. Similarly count all the possible cases, and make sure one particular person receives more than 2 books, in this case, person 3 should receive more than 2 books, always. Generally it is 6 ! p ! q ! r ! \sum\nolimits \frac{6!}{p!q!r!} where p + q + r = 6 p+q+r=6 , and r > 2 r>2 , where p , q , r p, q, r are non-negative integers.

Method 2: Same logic as method 1, just a simpler way to calculate the answer. It is the co-efficient of x 6 x^{6} in the following expression: 6 ! ( 1 + x 1 ! + x 2 2 ! + x 3 3 ! + . . . ) 2 ( x 3 3 ! + . . . . ) 6!(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...)^2(\frac{x^3}{3!}+....) . It is because, say, ( p , q , r ) (p,q,r) is a possible triplet. Thus, one of the terms in the co-efficient of x 6 x^6 is 6 ! p ! q ! r ! \frac{6!}{p!q!r!} , which is what is needed. Also notice that 1 + x 1 ! + x 2 2 ! + x 3 3 ! + . . . = e x 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...=e^x . Hence answer is co-efficient of x 6 x^{6} in 6 ! e 2 x ( e x 1 x x 2 2 ! ) = 6 ! ( e 3 x e 2 x x e 2 x x 2 e 2 x 2 ! ) 6!e^{2x}(e^{x}-1-x-\frac{x^2}{2!})=6!(e^{3x}-e^{2x}-xe^{2x}-\frac{x^2e^{2x}}{2!}) . Thus, the answer would be 6 ! ( 3 6 6 ! 2 6 6 ! 2 5 5 ! 2 4 4 ! 2 ! ) = 233 6!(\frac{3^6}{6!}-\frac{2^6}{6!}-\frac{2^5}{5!}-\frac{2^4}{4!2!})=233

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