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In how many ways can 10 balls be divided between two boys, one receiving two and the other eight balls
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The correct answer is:
$90$
$A$ gets $2, B$ gets $8 ; \frac{10!}{2!8!}=45$
$A$ gets $8, B$ gets $2 ; \frac{10!}{8!2!}=45$
$\therefore 45+45=90$.
$A$ gets $8, B$ gets $2 ; \frac{10!}{8!2!}=45$
$\therefore 45+45=90$.
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