In how many ways can 3 children divide among themselves 10 tickets chosen from 15 consecutively numbered tickets, such that each child receives a consecutive block of tickets — one child gets a block of 5, one gets a block of 3, and one gets a block of 2?

Question

Accepted Answer

C(8,5) * 3!. For any fixed ordering of the three block sizes along the number line, the gaps between and around them must sum to 5, giving C(8,3) = C(8,5) = 56 ways. Since there are 3! = 6 orderings of the block sizes, and assigning the resulting blocks to specific children is already captured by the 3! factor (each ordering naturally assigns a block to a child), the total is C(8,5) * 3!.

In how many ways can 3 children divide among themselves 10 tickets chosen from 15 consecutively numbered tickets, such that each child receives a consecutive block of tickets — one child gets a block of 5, one gets a block of 3, and one gets a block of 2?

Solution

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