Combinatorics-Self Doubt

Question

Number of ways to distribute 5 distinct toys to 3 children A,B,C such that each one of them gets atleast one toy.

My answer come to be 75.Is it correct?

 

Also, when toys are not distinct, will answer change?

Comments

@MiNiPanda I AM ALSO DOING LIKE THE SOLUTION BELOW

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BUT 144 NOT 150

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i'm get $147$

Answer 1

Each toy can be given to one of the three children: 3⁵

Subtract the number of ways atleast one child is not given a toy from total outcomes.

Using inclusion exclusion 3⁵ -(³C₁*2⁵ - ³C₂*1⁵) =150

Another approach

As @MiniPanda said the problem can be considered as how many onto functions are there from a set of 5 to a set of 3.

Comments

Okay. And when toys are not distinct, the answer is 75 right?

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If  toys are not distinct then,

WAY 1

c1+c2+c3=5

Since each child gets atleast one toy:

c1+c2+c3=2

Using stars and bars technique ( https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics))

^2+3-1C₂ =⁴C₂=6

OR

WAY 2

Child1	Child 2	Child 3
1	1	3
1	3	1
3	1	1
2	2	1
2	1	2
1	2	2

 

Answer 2

The answer is the Stirling number of the second kind.

Read more here:

https://math.stackexchange.com/questions/1495278/choosing-distinct-balls-to-put-into-indentical-urns
https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind