Question 360156
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If he has to work exactly 6 out of 10 and 3 out of the first 4, then he must work exactly 3 out of the last 6.  Hence, the answer to this question is to be found in the sum of the number of ways to select 3 things from 4 things plus the number of ways to select 3 things from 6 things:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4!}{3!(4\,-\,3)!}\ +\ \frac{6!}{3!(6\,-\,3)!}]


You can do your own arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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