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**: please read my approach and comment why it's wrong

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## Comments

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TopNewestYour 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.

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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)

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