Permutations (Combinatorics/Probability)

[BlakeH710]

Hi Instructors,

I've only recently started to review probability and combinatorics, so this may be a very ignorant question. Anyway, I came across a Kaplan Youtube video on the topic that offered some practice questions. The solution to one entialed using Permutation formula, which I don't believe is in any of the Manhattan Prep chapters on Combinatorics. Am I wrong? If not, is there a different method the Manhattan Prep suggests for combinatorics problems? The closest thing I've seen is perhaps the anagram grid or the glue method?


By the way, just for some context, the question was something along the lines of this:

"How many different pos. integers with 6 digits are there, when one digit is 5, one digit is 1, two of the digits are 2, and the remaining digits are a 6 or an 8."

Thanks,
Best Blake

[Sage Pearce-Higgins]

Thanks for your question. First of all, I should emphasize that combinatorics is not a big topic on GMAT: our research suggests that recent exams have usually been giving students only one combinatorics or probability problem per test.

In terms of the formula that you mention, I assume you're talking about the 'n choose r' formula. For example, if I have 8 people, then the number of 3-person teams I can choose is 8!/(5!3!). We effectively show this formula on page 61 of the number properties strategy guide (chapter 4 if you're using the e-book). However, we don't present it as a formula. Since most combinatorics problems are more complicated than the one I stated and contain the kind of twists that the problem you quoted does, having a generic formula doesn't help most students. For that reason, we recommend the grid method described in that same chapter. Let's apply it to your problem: "How many different pos. integers with 6 digits are there, when one digit is 5, one digit is 1, two of the digits are 2, and the remaining digits are a 6 or an 8."

I can write the information about the 6 digits in a table in no particular order:

5 1 2 2 6 8

I use this to create a fraction. Since there are 6 columns in my grid, the top of the fraction will be 6!. The bottom of the fraction will be 1!1!2!1!1! i.e. a factorial for each of the different "categories" in the 6 columns. This gives me a result of 6! / (1!1!2!1!1!) = 360