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$6$ boys and $8$ girls, how many ways to create a group of $4$ that has more than $1$ boy.

I know the solution of ${6\choose 1}{8 \choose 3} + {6\choose 2}{8\choose 2} + {6\choose 3}{8\choose 1} + {6\choose 4}{8\choose 0}$

but while I can't argue that this solution is wrong: ${6\choose 1}{13\choose 3}$, it seems to have overcounted.

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    If you want more than one boy you should not have the term $C(6,1)C(8,3)$ as that gets one boy and three girls. – 2017-01-23
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    @Kevin Long: Though I like the format you have used for combinations, I think we should stay with OP's format as that is what s/he is used to. The answers follow it. – 2017-01-23
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    @RossMillikan That's fair, I probably should've taken that into account. I'll leave it as is because I don't think OP will get confused by just changing his/her own answer, but if someone else deems it necessary, they can revert it. – 2017-01-24

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Suppose, in $C(6,1)$, say you select a boy $a$ and subsequently in $C(13,3)$ you select $3$ boys $(b,c,d)$

Suppose, in $C(6,1)$, say you select a boy $b$ and subsequently in $C(13,3)$ you select $3$ boys $(a,c,d)$

These are essentially same. This is just an example I have taken and many such cases exists which are counted more than one time.

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    awsome! it all makes sense now! – 2017-01-23
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Your $C(6,1)C(13,3)$ counts the ways of picking one boy and then picking three from the remaining people. There are two problems with this as an answer. First, it does not guarantee that there is more than one boy as the last three could be girls. Second, it overcounts the cases where there is more than one boy. If there are two boys, either one could be the one selected first, so you count this combination twice. If there are three boys you count it three times.