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Statistics problem (I think)?
You're taking an exam and you must answer EXACTLY 9 items out of 12. In how many ways can you select the 9 problems if:
a) you can choose ANY 9 out of 12 items
b) you must choose 4 from the first 6 and 5 from the last 6 problems
c) you must choose *at least* 4 from the first 6 problems.
It would be great if you guys could explain it.
4 Answers
- 1 decade agoFavorite Answer
These are just mildly fancy combinatorics problems. Do you remember learning about permutations and combinations in your probability course? This problem uses combinations since "order doesn't matter." Meaning: choosing problems #1 and #3 is the same as choosing #3 and #1. See?
I still haven't figured out how to type cool math symbols here. But, if you look at my second link, below, (on Combinations), you'll see some notation that has a little number, then a big "C", then another little number.
Looks sort of like.... nCk and is pronounced "n choose k"
Anyhow, that is the way to choose "k" items from a possible "n" items. Or, like in your part (a) of your question, to choose 9 items from a set of 12, you'd solve 12C9.
It's easy. Right? 12C9 = 12! / ( 3! * 9!) = 220. (You might notice this is the same as choosing 3 of the 12 problems you DON'T want to solve... 12C3 = 12C9 = 220.)
Part b just makes you break the exam up into two parts. You can combine the first choice with the second choice in any way, so you multiply the two "sub-answers."
Choose 4 from 6 (6C4) and then choose 5 from the last 6 (6C5) and then multiply them:
6C4 * 6C5
= 6! / (4! * 2!) * 6! / (5! * 1!)
= 15 * 6
= 90
The last one is a bit trickier. You have to do a problem like part (b) three times:
(i) choose 4 from the first 6 OR
(ii) choose 5 from the first 6 OR
(iii) choose all of the first 6 and then add these three together.
(i) 6C4 * 6C5 = 90
(ii) 6C5 * 6C4 = 90
(iii) 6C6 * 6C3 = 20
SUM = 200.
Recap:
a) 220
b) 90
c) 200
- 1 decade ago
a)
getting 9 out of 12 equals leaving 3 out of 12
12 ways to leave 1st question, 11 for second and 10 for third
= 12*11*10 = 1320 ways.
b)
Break the problem into 2 parts ( 4 out of first 6 and 5 out of last 6)
A : choosing 4 out of first 6 as as good as leaving 2 out of first 6
= 5*6 = 30 ways
B: Choosing 5 out of 6 is as good as leaving one = 6 ways.
Total = A * B = 30*6 = 180 ways.
c)
Choosing exactly 4 from first 6 will be like case b = 30 ways
Choosing 5 from first 6 will be 6 ways * (choosing 4 from remaining 6) = 6 * 30 = 180
Choosing 6 out of first six will be 1 way * (ways to choose remaining 3 from 6) = 1*(6*5*4) = 120
Hence Total ways = 30 + 180 + 120 = 330
- Mugen is StrongLv 71 decade ago
qa
answer
= 12C9
= 12!/9!3!
= 220
qb
answer
= 6C4 * 6C5
= 6!/4!2! * 6!/5!1!
= 6*5*6/1*2*1
= 90
qc
answer
= 6C4*6C5 + 6C5*6C4 + 6C6*6C3
= 2*90 + 1*20
= 200
- 1 decade ago
a. 12*11*10*9*8*7*6*5*4*3
You have 12 choices for the first problem 11 for the second..and so on.
b. 6*5*4*3*6*5*4*3*2
c.6*5*4*3*8*7*6*5*3
you're basically multiplying the choices you have.
in b. you have 6 for the first, 5 for the second but after the fourth, you must do 5 of the last 6 so you have again 6 choices.