The University of Maryland, University of Vermont, and Emory University have each $4$ soccer players. If a team of $9$ is to be formed with an equal number of players from each university, how many number of ways can the selection be done?

Possible answers are: $3, 4, 12, 16, 25$.

I have encountered that problem in GRE Quant.

Here is my solution: From each university we should take $9:3=3$ players and this choice could be done with $C_4^3=4$ ways. Hence the selection can be done in $4\times 4 \times 4=4^3=64$ ways. But such answer does not exist.

Can anyone explain this moment please?

  • 1
    $\begingroup$ I think this must be a mistake? Your reasoning is correct. $\endgroup$ – Fimpellizieri Aug 25 '17 at 5:13
  • $\begingroup$ 64 is correct , can you recheck this question for any missing words $\endgroup$ – Samar Imam Zaidi Aug 25 '17 at 5:20
  • $\begingroup$ Does the ungrammatical phrase "how many number of ways" actually appear in the original text? $\endgroup$ – Eric Towers Aug 25 '17 at 12:46
  • $\begingroup$ @EricTowers, Yes it does! $\endgroup$ – ZFR Aug 26 '17 at 6:30
  • $\begingroup$ When this question was asked three years ago, math.stackexchange.com/questions/843152/… , the answer was also $64$. $\endgroup$ – Eric Towers Aug 26 '17 at 19:25

I cross check with other answer is 12=4+4+4

The question is asking selection for each university hence Principal of Addition will apply. Principal of Multiplication will be applied when asked about number of possible combination of team. The question specifically asks selection from each university not selection of team

  • $\begingroup$ Can you explicitly illustrate these 12 selections? $\endgroup$ – ZFR Aug 25 '17 at 5:38
  • $\begingroup$ This question is tricky and test Principal of Addition/Multiplucation Concept, GRE has purposefully removed 64 so that people will ponder over this question $\endgroup$ – Samar Imam Zaidi Aug 25 '17 at 5:40
  • $\begingroup$ The answer 12 is not persuasive. I would like to look at these 12 choices in some certain example. $\endgroup$ – ZFR Aug 25 '17 at 5:41
  • $\begingroup$ It is testing "Principal of Addition" vrs "Principal of Multiplication" concept. Number of Sekection from Each university=4 total selection =12, but if asked number of team formed it is 4*4*4=64 $\endgroup$ – Samar Imam Zaidi Aug 25 '17 at 5:45
  • $\begingroup$ Still can not understand you. Could you explain it in real example? $\endgroup$ – ZFR Aug 25 '17 at 5:49

I'd frame exactly the same computation you use as "the number of ways to pick the one player that is excluded from each University ...". Of course, I'd get the same answer...

As a quick enumeration of at least 26 different ways to make some of these teams...

Let $a, b, c, d$ represent the four players from the University of Maryland. Let $f, g, h, i$ represent the four players from the University of Vermont. Let $p, q, r, s$ represent the four players from Emory University. The question is this ungrammatical mess "If a team of 9 is to be formed with an equal number of players from each university, how many number of ways can the selection be done?" In particular, "how many number of ways" is not grammatically correct English. Nevertheless, we will represent a "selection" by who from each University is not selected (since listing the three selected players is equivalent to listing the one non-selected player from each University). \begin{align*} afp && afq && afr && afs \\ agp && agq && agr && ags \\ ahp && ahq && ahr && ahs \\ aip && aiq && air && ais \\ \\ bfp && bfq && bfr && bfs \\ bgp && bgq && bgr && bgs \\ bhp && bhq && &&\text{+2 more} \\ &&&&\text{+4 more} \\ \\ &&&&\text{+16 more} \\ \\ &&&&\text{+16 more} \\ \\ \end{align*}

  • $\begingroup$ I did not get you. Do you mean that my answer is correct? $\endgroup$ – ZFR Aug 25 '17 at 5:25
  • $\begingroup$ @RFZ : Yes. I also get $64$. $\endgroup$ – Eric Towers Aug 25 '17 at 5:27
  • $\begingroup$ Hmm. Such problems with weird answer options in GRE official books seem very strange $\endgroup$ – ZFR Aug 25 '17 at 5:36

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