Here I went straight into permutations/combinations, if this were the case it would be a combination because order does not matter, however this is not a permutation/combination problem. My question is how do I tell the difference of what method to use, here they just took the 5 and multiplied by 8 5's, how would I know to do this instead of the permutation/combination formulas?
Hey Diego,
I guess the main thing to think about is that when you're picking a combination or permutation, you're picking a collection of items from ONE SINGLE GROUP of whatever size, and you're not returning an item to that group once you've picked it to potentially be picked again (e.g. if I am picking a combination of four shirts to pack from seven I own, then once I pick the yellow one I don't then return it to the pile to potentially pick it on my next draw too). This is a different scenario from the test question here. The test question here you can either think of as there being 8 DIFFERENT GROUPS (of five apiece), unrelated to each other, in which case once you make your selection from one group you move onto a whole new group to choose from... or you can think of it as just one group of five answer choices (say A through E), but even if so, you're replacing whatever answer choice you've drawn after each draw, unlike in a combination.
If you think of it the first way, it's not WHOLLY unrelated to combinations, in that you could think of it as that for EACH question, you have to choose 1 from 5 possibilities, so if each question yields 5C1, i.e. 5!/((5-1)!1!), i.e. 5 possible combinations, then we multiply all those 5s together in the string of 8 of them we wind up with. But it's obviously not necessary to use the combination formula when you're just choosing 1 from something, because it's easy to see how many choices you have in that case. You wouldn't go wrong using it, though. The thing is just that you wind up with those eight separate combinations, because it's eight separate groups.
If you think of it the second way (one group, with replacement of an answer choice after each time it's picked), then we're not dealing with a combination in any way, because we're never setting aside items from the group. Do think for contrast, though, about what this would look like as a permutation. For instance, if the setup of the question were that this were one of those matching quizzes, where you've got a column of letters and a column of numbered things and you have to match the appropriate letter with the appropriate number. Say we only had the letters A through E and we had 8 numbered questions, and we were supposed to put a letter next to the numbered question it corresponded to, but we could only use each letter once (and three of the questions would get no response). As an example, if I had a matching quiz like this:
A. 1215 ____ 1. Columbus' first voyage
B. 1492 ____ 2. Soviet Union dissolved
C. 1776 ____ 3. Brown vs. the Board
D. 1863 ____ 4. Magna Carta
E. 1954 ____ 5. Bombing of Hiroshima
____ 6. Gettysburg Address
____ 7. Declaration of Independence
____ 8. Bay of Pigs
THEN we'd have a permutation -- 8 P 5 -- 8!/(8-5)! or in other words (8)(7)(6)(5)(4) -- ways I could possibly fill out this quiz -- because I am choosing five slots from one group of 8, AND order is relevant, since for any given five I choose, it's going to be a different outcome each time I shuffle the order of the letters I put in them.
Hope this helps,
Ashley
Hey Diego,
I guess the main thing to think about is that when you're picking a combination or permutation, you're picking a collection of items from ONE SINGLE GROUP of whatever size, and you're not returning an item to that group once you've picked it to potentially be picked again (e.g. if I am picking a combination of four shirts to pack from seven I own, then once I pick the yellow one I don't then return it to the pile to potentially pick it on my next draw too). This is a different scenario from the test question here. The test question here you can either think of as there being 8 DIFFERENT GROUPS (of five apiece), unrelated to each other, in which case once you make your selection from one group you move onto a whole new group to choose from... or you can think of it as just one group of five answer choices (say A through E), but even if so, you're replacing whatever answer choice you've drawn after each draw, unlike in a combination.
If you think of it the first way, it's not WHOLLY unrelated to combinations, in that you could think of it as that for EACH question, you have to choose 1 from 5 possibilities, so if each question yields 5C1, i.e. 5!/((5-1)!1!), i.e. 5 possible combinations, then we multiply all those 5s together in the string of 8 of them we wind up with. But it's obviously not necessary to use the combination formula when you're just choosing 1 from something, because it's easy to see how many choices you have in that case. You wouldn't go wrong using it, though. The thing is just that you wind up with those eight separate combinations, because it's eight separate groups.
If you think of it the second way (one group, with replacement of an answer choice after each time it's picked), then we're not dealing with a combination in any way, because we're never setting aside items from the group. Do think for contrast, though, about what this would look like as a permutation. For instance, if the setup of the question were that this were one of those matching quizzes, where you've got a column of letters and a column of numbered things and you have to match the appropriate letter with the appropriate number. Say we only had the letters A through E and we had 8 numbered questions, and we were supposed to put a letter next to the numbered question it corresponded to, but we could only use each letter once (and three of the questions would get no response). As an example, if I had a matching quiz like this:
A. 1215 ____ 1. Columbus' first voyage
B. 1492 ____ 2. Soviet Union dissolved
C. 1776 ____ 3. Brown vs. the Board
D. 1863 ____ 4. Magna Carta
E. 1954 ____ 5. Bombing of Hiroshima
____ 6. Gettysburg Address
____ 7. Declaration of Independence
____ 8. Bay of Pigs
THEN we'd have a permutation -- 8 P 5 -- 8!/(8-5)! or in other words (8)(7)(6)(5)(4) -- ways I could possibly fill out this quiz -- because I am choosing five slots from one group of 8, AND order is relevant, since for any given five I choose, it's going to be a different outcome each time I shuffle the order of the letters I put in them.
Hope this helps,
Ashley
