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10 different toys are to be distributed ...

10 different toys are to be distributed among 10 children. Total number of ways of distributing these toys so that exactly 2 children do not get any toy, is equal to:

A

`(10!)^(2)((1)/(3!2!7!)+(1)/((2!)^(5)6!))`

B

`(10!)^(2)((1)/(3!2!7!)+(1)/((2!)^(4)6!))`

C

`(10!)^(2)((1)/(3!7!)+(1)/((2!)^(5)6!))`

D

`(10!)^(2)((1)/(3!7!)+(1)/((2!)^(4)6!))`

Text Solution

Verified by Experts

The correct Answer is:
B
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