From $100$ used cars sitting on a lot, $20$ are to be selected for a test designed to check safety requirements. These $20$ cars will be returned to the lot, and again $20$ will be selected for testing for emission standards.

a) In how many ways can the cars be selected for both tests if exactly $5$ cars must be tested for safety and emission.

The answer that is listed in the book for this question is $$\binom{100}{5}\binom{95}{15}\binom{80}{15}$$

I'm slightly confused with this answer. The answer I came up with at first was: $$\binom{100}{20}\binom{20}{5}\binom{95}{15}$$ This made sense to me because first you choose $20$ cars for safety. Then you need to choose $5$ out of those $20$ for emissions, and finally you choose $15$ other cars out of $95$ for emissions. Why does this not work?

  • $\begingroup$ Okay thanks, I changed it to make it a bit clearer $\endgroup$ – bugsyb Nov 8 '15 at 23:35
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Nov 9 '15 at 11:01

You don't want to choose $15$ from $95$ for emissions testing, since the $20$ chosen in your $\binom{100}{20}$ are not eligible. So your last term should be $\binom{80}{15}$.

After you make that change, your expression will look different from the given answer, but it turns out to be the same.There are often several ways to solve a combinatorial problem.


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