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

#Combinatorics

Easy Math Editor

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Comments

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.

Mark C - 5 years, 1 month 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)

D K - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$