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by sana.noor » Sat Jul 27, 2013 9:57 am
Q: There are 21 identical Ts and 19 identical Es. In howmany ways can these items can be placed in a Shelf such that no two Es are together?

A: 1540
source: btg forum
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by Matt@VeritasPrep » Sat Jul 27, 2013 11:11 am
Let's illustrate with smaller #'s first.

Say I have 4 T's and 2 E's, with the same restrictions. I can't have two E's next to each other, so I essentially have this shelf:

_ T _ T _ T _ T _

Where each of the _'s represents a space in which I can place one E.

I have two E's and five spaces, so the number of arrangements is (5 choose 2), or 5!/(2! * 3!). Once I choose the spaces, I'm done - no more arranging.

Since we have 21 T's, we have 22 spaces into which we place our E's. We have 19 E's, so our equation is (22 choose 19), which gives 1540.
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by Brent@GMATPrepNow » Sat Jul 27, 2013 2:36 pm
For additional practice with this topic, here's a similar question: https://www.beatthegmat.com/combinations-t123264.html

I have one small thing to add to Matt's great solution. Once we know that we need to calculate 22C19 (22 choose 19) to find the answer, we can use the fact that 22C19 = 22C3 and then calculate 22C3 in our heads, using the technique shown in this free video: https://www.gmatprepnow.com/module/gmat-counting?id=789

Aside: In general, nCr = nC(n-r) . . . or we could say "n choose r" is equal to "n choose n - r"
So, for example: 10C7 = 10C3
12C9 = 12C3
8C7 = 8C1
etc.

Cheers,
Brent
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