<p>You can choose 1 main disk, 1 drink, 1 dessert from 3 main dishes, 4 drinks and 5 desserts. How many combination you can get?
a) 12
b)24
c)60
D)120</p>
<p>3<em>4</em>5 = 60 -> (c)</p>
<p>That's one of the easiest Q's you'll come across. Simply multiply the values</p>
<p>You understand why right? There are three choices for the main dish, four choices for the drink, and five choices for the dessert. So you multiply them together to get the total number of choices.</p>
<p>^Expanding on that, and taking just the dish and drinks first, for each dish, you could have 4 drinks. So for 1st dish, you could have 4 possible drinks, for the second dish you could have 4, and for the 3rd dish, you could have 4. So there are 3 dishes * 4 drinks = 12 combinations of a dish and a drink. </p>
<p>The same principle holds when you factor in the desserts. For each of the 12 dish/drink combinations, you could have 5 desserts. (For the first one, you could have 5, for the second one you could have 5, ...) 12*5=60.</p>
<p>Why not add them up but multiply? I mean, why not any other but multiplication?</p>
<p>sreis explained it conceptually pretty well. If that doesn't make sense, lets just take an example. From one group I can choose A, B, or C. From another I can choose D, E, or F. </p>
<p>If I want to take one from each group, I can take AD, AE, AF, BD, BE, BF, CD, CE, or CF - 9 options. </p>
<p>3 x 3 = 9, (right)
3 + 3 = 6, (wrong)
3 / 3 = 1, (wrong)
3 - 0 = 0. (wrong)</p>
<p>Multiplication works.</p>
<p>Hmm so for combination always multiply after you figured out the answer for the choices whether it's 2 out of 5.. etc.. Cuz i can do that on the calculator so i just needed to know to multiply the values for all combinations problems.</p>
<p>Is this the same with permutation usually too?</p>
<p>It's not the same. For those who don't know, in permutations order matters, meaning that a different order implies a different permutation: {steak,soda,pie} ≠ {soda,steak,pie}. You count them both, so there are two permutations. In combinations they are equivalent, so there is just one combination there. </p>
<p>If you wanted to know the permutations of the above, you just need to know how many ways there are of ordering 3 elements, and it's 3!. In other words, you can order each of the combinations above in 6 different ways. 60*6=360.</p>
<p>so finding permutations of multi-faceted stuff where the traditional 5! doesn't work.. you multiply the number of combinations by the number of possible slots? or can you explain.</p>
<p>Yeah, well I'm not sure where 5! came from. Do you mean n!, as in the traditional, plain factorial? Or is 5! supposed to refer to the problem?</p>
<p>I used 3! because there are three items that could vary in their order. You have to do this after the combinations because you find the permutations of a set (order doesn't matter in a set). You need a set first. This is only important if you know order matters.</p>
<p>I'd guess that in most SAT permutation problems you would <em>just</em> do a permutation. For example, if there are three people in a car, and each person could either be a driver, front passenger, or back passenger, how many ways could they sit in the car? </p>
<p>If the people are A, B, and C, then there is just one set {A,B,C}. For sets {A,B,C} = {B,A,C}. For permutations ABC ≠ BAC. Permutations go after combinations in the original problem because there were 60 distinct sets ("distinct" is redundant). The only method I could say is to use combinations when order doesn't matter and permutations when it does. If you want a better way, maybe check out a chapter on counting in a math book. Anyways, I wouldn't worry too much about this problem because it's unlikely you would be asked to find both combos and permutations.</p>
<p>1) Why in this math problem we can't use sth like x!/ n!(x-n)!
2) What if the problem change to: 2 dishes, 1 drink, 1 desert?</p>
<p>1) x!/ n!(x-n)! is when you are trying to find the number of distinct sets of size n or (x-n) from x items. You aren't doing that here.
2) 2<em>1</em>1. If you have 1 drink and 1 dessert, you have no choice about them. All you are deciding is to have dish1 or dish2.</p>
<p>x!/ n!(x-n)! ~ you don't need to know this formula for SAT1</p>
<p>A lot of SAT prep books present it.
but we're trying to find the distinct set of dish and drink etc in this problem, aren't we? So I still don't get why we can't use that formula</p>
<p>The formula finds distinct sets of a certain size from a larger set. For example you might try to find say all distinct sets of size 3 from 5 elements, so it would be 5!/(3!*2!). In the problem you aren't trying to break up a larger set into smaller sets.</p>
<p>Is the problem similar to this:
We have 40 senior who is competing for their class officers. Only 5 of them will be picked. How many possible way will there be?
W/ this we use that formula above right?
So my thinking to the original problem is that:
We have (3) drink, how many possible way can we pick 2 drinks out of it?
In the same manner, we eventually got the result of (permutation of each of the three chances) and that's I think the answer.
But it's wrong.
Well, can you show me where I went wrong in my thinking? Thanks a lot</p>
<p>Yeah it was ambiguous...</p>
<p>When you said change to "What if the problem change to: 2 dishes, 1 drink, 1 desert?" there were two things it could be because the problem was "You can choose 1 main disk, 1 drink, 1 dessert from 3 main dishes, 4 drinks and 5 desserts." I didn't know if you were referring to the left part or right part. </p>
<p>But anyway, there's more ambiguity. If you can't choose the same dish twice, then yes its 3!(2!*1!)=3, but if you can choose the same main dish twice and order doesn't matter then it's 6 because you add the duplicate-type choices.</p>
<p>Oh I'm sorry I'm supposed to say the first problem. it's 1, not 2. But still, the formula doesn't work.
Sreis, how do you come up with 3!(2!*1!)=3? I thought the formula is x!/ x!(x-n)!</p>
<p>sorry 3!(2!<em>1!)=3 should read 3!/(2!</em>1!)=3, where there is a / after the 3!.</p>