<PRE><B>Question 653377</B>
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I don&#39;t know where you got the idea that you needed to use permutations -- If the two boys are Bobby and Billy that would be the same as if they were Billy and Bobby, right?


You need the number of &lt;b&gt;&lt;i&gt;combinations&lt;/i&gt;&lt;/b&gt; of 6 boys taken 2 (NOT 5) at a time times the number of combinations of 4 girls taken 3 at a time.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ {{6}\choose{2}}\ \times\ {{4}\choose{3}}\ =\ \frac{6!}{4!2!}\ \times\ \frac{4!}{3!1!}\ =\ \frac{6\times5}{2}\ \times\ 4]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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