In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy? 
  
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*In how many ways can we split $6$ toys such that John will receive $2$, Carl and Michael will receive at least one toy?
  

I'll be using combinations in this case. Let us see how many ways there are to split $6$ boys as seen below
$$\binom{6}{2}$$
This represents that John will receive any $2$ toys among $6$ toys. From remaining toys, we can conclude that
$$\binom{4}{1}\binom{3}{1}$$
We've found the case that Michael and Carl receives only one toy. Now we have to consider that they will receive $2$ toys since it says at least. 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
Finally, I'm combining all these results and we have that 
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{6}{2}\binom{4}{1}\binom{3}{1} = 90 + 180 = 270$$
I've gone wrong somewhere. Could you assist me with that?
 A: John get 2, that is ${6\choose 2}$. 
Now from the rest of 4 Carl gets: 
$1$ toy, that is ${4\choose 1}$ way or
$3$ toys, that is ${4\choose 3}$ ways or 
$2$ toys, that is  ${4\choose 2}$ ways. So the number of all distributions is   $${6\choose 2}\Big( {4\choose 1}+ {4\choose 2} + {4\choose 3}\Big)$$. 
A: Since John receives two toys and Carl and Michael each receive at least one toy, the possible distributions are $(2, 1, 3)$, $(2, 2, 2)$, and $(2, 1, 3)$, where the ordered triple $(J, C, M)$ represents the number of toys received by John, Carl, and Michael, respectively.  
If John receives $J = 2$ of the six toys and Carl receives exactly $C$ of the remaining four toys, then Michael will receive all of the remaining $M = 4 - C$ toys.  Hence, the number of ways of distributing the toys so that John receives two of the six toys and John and Carl each receive at least one is 
$$\binom{6}{2}\binom{4}{1}\binom{3}{3} + \binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{6}{2}\binom{4}{3}\binom{1}{1}$$
