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k_nkavoyannis
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A men's basketball league assigns every player a two-digit

by k_nkavoyannis Wed Dec 16, 2009 1:53 pm

I am having a general problem getting my head around the basic principles of combinations and permutations. I can show you what I mean with the following question, which is #3 on page 75 of the 4th Edition of the Word Problems Strategy Guide.

"A men's basketball league assigns every player a two-digit number for the back of his jersey. If the league uses only the digits 1-5, what is the maximum number of players that can join the league such that no player has a number with a repeated digit (e.g. 22), and no two players have the same number?".

I solved this using the formula for permutations: 5!/(5-2)! = 5!/3! = 5x4 = 20. This was of course the right answer.

However, what would I have done if there had been no restriction on doubled numbers, i.e. 11 and 22 and so on were allowed? I would still have used the same formula and then later realized that of course the answer in this case would be 25, not 20, so the formula doesn't apply. In more complicated questions though, under the pressure of time, I would make the mistake.
k_nkavoyannis
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Re: A men's basketball league assigns every player a two-digit

by k_nkavoyannis Sat Jan 09, 2010 8:36 am

Thanks
Ben Ku
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Re: A men's basketball league assigns every player a two-digit

by Ben Ku Thu Jan 21, 2010 3:20 am

When you use the permutation formula, you are selecting DIFFERENT values from your set. For example, here, 5P2 means we are selecting two different numbers from the five possible. The reason why this works for you in this problem is because the two digits are related to each other: the second digit cannot be the same as the first digit.

(The recommended approach would be to use the slot method: "there are five options for the first digit and four options for the second digit, so 5*4 = 20")

However, if we remove the restriction of no repeated digits, then basically the two digits are not related to each other at all. In this case, the first digit would be 5C1, and the second digit would also be 5C1, so 5C1 * 5C1 = 5 * 5 = 25.

Of course this is the hard way of doing it. You could just say "there are five options for the first digit and five options for the second digit, so using the slot method, 5 * 5 = 25."

I would try using the SLOT method first, then using the ANAGRAM approach second. Use formulas only if you know how they work.
Ben Ku
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ManhattanGMAT