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Question

10 different toys are to be distributed among 10 children. Total number of ways of distributing these toys, so that exactly two children do not get any toy, is

A
10!2!3!7!
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B
10!(2!)4×6!
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C
(10!)2[1(2!)4×6!+12!×3!]
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D
10!×10!(2!)2×6![2584]
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Solution

The correct option is D 10!×10!(2!)2×6![2584]
There may be two cases.
Case I. Two children get none and one get three and others get one each.
Then, total number of ways =10!2!×3!×7!×10!
Case II. Two get none and two get 2 each and the other get one each.
Then, total number of ways =10!(2!)4×6!×10!
Hence, total number of ways
=10!2!×3!×7!+(10!)2(2!)4×6!=(10!)2×25(2!)2×6!×84.

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