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is there any other approach to solving this problem?
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abheeeee
is there any other approach to solving this problem?

nCk = nC(n - k) is the easiest to compute for k = 2 being equal to \(\frac{n(n-1)}{2}.\)
You should start with the highest values and go backwards, as 6C3 = 20 and nCk is the highest for k around half of n.
So, you can deduce that n must be greater than 6.
And take advantage of the multiple choice question, as there is just one correct answer.

Otherwise, once you understand Pascal's triangle, it is quite easy to write it down. Having all the values in front of yours eyes, really helps to pinpoint the correct answer.
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For choosing teams, you want to keep in mind the combination basics...
you want to choose out of n people, create a team of r. THe number of combinations (NOT permutations) should be between 21-24. That's more than 20 and less than 25.

So, if you try the various combinations the only one that works is out of 7 choose a team of 5.

THat's 7NCR5
7! / (5! 2!) = 6*7/2 = 21

What about 8C4?

8! / (4!*4!) = 5*6*7*8/4 = 30*7*2 = 240

What about 8C5?
8! / (5! 3!) = 6*7*8 / (3*2) = 56

What about 8C6?
8! / (6! 2!) = 7*8 / 2 = 28

What about 8C7?
8! / (7! 1!) = 8

So somehow we don't get in the correct range. You can try for the others but you won't get in range.

6C3 = 6! / (3! 3!) = 4*5*6 / (3*2) = 20
6C4 = 6! / (4! 2!) = 5*6 / 2 = 15

The highest number of combinations will be in the middle. So 6C3 will be the max for 6 people. 4C2 will be the max for 4 people. So if 6C3 only has 20, we know we need to go higher to 7.

7C4 = 7! / (4! 3!) = 5*6*7 / (3*2) = 5*7 = 35 (too high)
7C5 = 7! / (5! 2) = 6*7 / 2 = 21 (just right!)

So if you're familiar with the binomial distribution curve for these combinations and that out of 8 you choose a number in the middle you'll get the highest number of results. Using that you can do an educated guess as to what to try next.
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whats the quicker way to do this? I started with 8C4 and kept checking until I found a value between 21 and 25. Bunuel
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Robo_123
whats the quicker way to do this? I started with 8C4 and kept checking until I found a value between 21 and 25. Bunuel

We need more than 20 but less than 25, with at least as many chosen as not chosen.

Start with 5 candidates. The maximum possible is 5C3, which equals 10, so that’s too small.

Next, try 6 candidates. Since we are told that at least as many are chosen as not chosen, we should check both 6C3 and 6C4. 6C3 is 20, which is not more than 20, and 6C4 is 15, which is less. So 6 does not work.

Now try 7 candidates. First, 7C4 gives 35, which is too high. Then check 7C5, which gives 21, fitting perfectly between 20 and 25.

So, 7 candidates and 5 players meet all conditions, and there’s no need to check further.
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I thought it says the combination of chosen team. Means if a team is chosen how many different combinations we can make. I dont know where does it says combination from the team is choosen.
GMATPill
For choosing teams, you want to keep in mind the combination basics...
you want to choose out of n people, create a team of r. THe number of combinations (NOT permutations) should be between 21-24. That's more than 20 and less than 25.

So, if you try the various combinations the only one that works is out of 7 choose a team of 5.

THat's 7NCR5
7! / (5! 2!) = 6*7/2 = 21

What about 8C4?

8! / (4!*4!) = 5*6*7*8/4 = 30*7*2 = 240

What about 8C5?
8! / (5! 3!) = 6*7*8 / (3*2) = 56

What about 8C6?
8! / (6! 2!) = 7*8 / 2 = 28

What about 8C7?
8! / (7! 1!) = 8

So somehow we don't get in the correct range. You can try for the others but you won't get in range.

6C3 = 6! / (3! 3!) = 4*5*6 / (3*2) = 20
6C4 = 6! / (4! 2!) = 5*6 / 2 = 15

The highest number of combinations will be in the middle. So 6C3 will be the max for 6 people. 4C2 will be the max for 4 people. So if 6C3 only has 20, we know we need to go higher to 7.

7C4 = 7! / (4! 3!) = 5*6*7 / (3*2) = 5*7 = 35 (too high)
7C5 = 7! / (5! 2) = 6*7 / 2 = 21 (just right!)

So if you're familiar with the binomial distribution curve for these combinations and that out of 8 you choose a number in the middle you'll get the highest number of results. Using that you can do an educated guess as to what to try next.
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manish8242
I thought it says the combination of chosen team. Means if a team is chosen how many different combinations we can make. I dont know where does it says combination from the team is choosen.

You're reading it as if the team is already chosen, but the phrase "distinct possibilities for the composition of the chosen team" means the coach is still choosing the team. It refers to how many different possible teams the coach can form from the pool, not what happens after one team is picked.
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