The number of ways of distributing six different books amongst three persons such that a particular person gets three or more books is:
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If person 1 is to get books 1, 2, 3, then he will receive it only 1 way and not 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 ! 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 2 ! 1 ! 3 ! 6 ! 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 ∑ p ! q ! r ! 6 ! where p + q + r = 6 , and r > 2 , where 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 in the following expression: 6 ! ( 1 + 1 ! x + 2 ! x 2 + 3 ! x 3 + . . . ) 2 ( 3 ! x 3 + . . . . ) . It is because, say, ( p , q , r ) is a possible triplet. Thus, one of the terms in the co-efficient of x 6 is p ! q ! r ! 6 ! , which is what is needed. Also notice that 1 + 1 ! x + 2 ! x 2 + 3 ! x 3 + . . . = e x . Hence answer is co-efficient of x 6 in 6 ! e 2 x ( e x − 1 − x − 2 ! x 2 ) = 6 ! ( e 3 x − e 2 x − x e 2 x − 2 ! x 2 e 2 x ) . Thus, the answer would be 6 ! ( 6 ! 3 6 − 6 ! 2 6 − 5 ! 2 5 − 4 ! 2 ! 2 4 ) = 2 3 3