Ugh, toughie. Actually, tough to explain. I'll try to give as many examples as possible.
1. P = (m + n)!/(m - 1)! will be P = m*(m + 1)*(m + 2)*...*(m + n). As you can probably guess, there are (n + 1) elements in this product (which I've noted as P for brevity). If you have trouble understanding this, let me give you an example:
(3 + 5)!/(3 - 1)! = 8!/2! = 3*4*5*6*7*8
n = 5 and you end up with 6 CONSECUTIVE elements in the product. Note that the first element in the product will always be m. In the above case, m = 3.
For P to be divisible by 16, your best bet is to see that 16 = 2^4, so you need at least 4 2's in your product. Now comes the hard part, which I hope I can explain to you in such a way that makes sense.
Every two numbers, you get an even number, i.e. 1, 2, 3, 4, 5, 6. So you'd think you need four even numbers for your product, right? Actually, you'll only need three at most. This happens because EVERY TWO EVEN NUMBERS, one of these two even numbers is divisible by 4 (this makes sense if you write a list of numbers as I did above with the even numbers). So you'll need three even numbers tops, since they'd all be divisible by 2 (three times 2), but one of the three will always be divisible by 4 (another 2).
Look at the example above. It's basically the worst case scenario, when the string of numbers starts with an odd integers. You need all 6 numbers to get the "at least four 2's", so n = 6 - 1 = 5. There are other more favorable cases, i.e. if you start with an even number:
4*5*6*7*8 - in this case you get two 2's from 4, one 2 from 6 and three 2's from 8. As you can see, here you only need 5 elements to get the "at least four 2's". Since the number of elements = n + 1, in this example n = 5 - 1 = 4. In this case, m = 4 (since it's the first element of the product).
BUT
Since n is devised in such a way that no matter what m is, P will be divisible by 16, n must be AT LEAST 5 (since 5 is our worst case scenario). However, as you might have guessed, there are more than one possible values for n, since n >= 5. As long as you get 6 numbers in your product, everything's cool, no matter m. So since 1 yields more than one value for n (actually, an infinity: every integer greater than or equal to 5), 1 is insufficient.
2. n^2 - 9n + 20 = (n - 5)(n - 4) = 0. This quadratic has two solutions: n = 4 and n = 5. Since you get two options, 2 is insufficient as well.
But put both statements together to get:
1. n > = 5
2. n = 4 or n = 5
As you can see, there's only one value that fits and that's 5. Since you get a single value for n, C is the correct answer.
Man I sure hope this makes sense to you. I know 1 seems pretty convoluted, but trust me: it's the examples that make it big. If you have enough practice to know what to look for, then I guarantee it's easier.
DS: For every integer m, (m+n)!/(m – 1)!
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Source: Beat The GMAT — Data Sufficiency |
