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How many ways are there to distribute 5 distinct toys among 3 children such that every child gets at least 1 toy?

but I’m getting 9.

My approach go as follows:

step-1 : give 1 toy to all 3 children, now, I am left with (5-3) = 2 toys. Therefore, now I have to distribute 2 toys among 3 children.

step-2 : for toy1, I can give it to first child or second child or third child. Similarly, I can give toy2 to either of first or second or third child.

thus, toy1 has 3 choices and so do toy2. Therefore, my answer would be 3*3  = 9.

You're approach is wrong, as we've distinguishable toys.

The answer to this problem is same as number of $onto$ functions from set of toys to set of children.
Try using cases :-

5 toys you have to distributed at least 1 toy to each so

there are two cases :-

distribute it as (2,2,1) or (3,1,1) .

total number of ways when you distribute is to (2,2,1) =$\binom{3}{2}*\frac{5!}{2!*2!}=90$

Total number of ways when you distribute it to (3,1,1)=$\binom{3}{1}*\frac{5!}{3!}=60$

So total number of ways is $(90+60)=150$ .

U will be missing a lot of cases because the toys are not identical to begin with