# Distribution of Toys

10 different toys are to be distributed among 10 children in a way that exactly 2 children don't get a toy. Total number of ways would be?

Note by D K
2 years, 3 months ago

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Your approach double-counts situations where a child gets multiple toys. Suppose she gets toys A and D; she can get A as one of the eight and D as one of the two, or vice versa; you count those as two different "ways" when they are equivalent.

- 2 years, 3 months ago

This is my approach( which gives the wrong answer):
First choose the 2 left outs- number of ways={10 \choose 2} ---(1)
Number of ways for 8 kids to get one toy each from 10 toys= 10X9X8...... 8-terms = {10 \choose 8} X 8! ----(2)
Number of ways to distribute remaining two toys= 8^2 -----(3)
Therefore, total number of ways to do the business= (1) X (2) X (3)

- 2 years, 3 months ago