Question 639088
On a table, there are 50 good pens and 10 defective pens.  In how many ways can 25 pens be selected so that exactly 3 are defective?
<pre>
"50 choose 22" TIMES "10 choose 3" = C(50,22)×C(10,3) = 1.065 × 10<sup>16</sup>

Edwin</pre>