A college plans to interview 8 students for graduate assistantships. The college has three assistantships available. How many groups of three can the college select?
2006-09-28 19:07:40 · 9 answers · asked by KISMET 2 in Science & Mathematics ➔ Mathematics
Sub population problem, so:
n = total number of students
r = number of students to be selected
C = n!/[r!*(n-r)!] = 8!/(3!*5!)
This comes out to be 56 possible sub-population.
To understand this better, think first in terms of ordered populations. If the order was important, you would have 8*7*6 = 336, or n!/(n-r)!. However, because you can have only one of each permutation, and each permutation comes in 3*2*1 different combinations, you actually have 1/6 of 336 sub-populations which is equal to 56.
2006-09-28 19:36:37 · answer #1 · answered by rmtzlr 2 · 2⤊ 0⤋
1
2006-09-29 02:58:22 · answer #2 · answered by Anonymous · 0⤊ 2⤋
I get 112.
Call the A, B, C, D, E, F, G, H
A can be paired with 7 other individuals and then the 3rd position has 6 possibilities
A 7 * 6 = 42
A is now out. B can be paired with 6 other individuals and the 3rd position has 5 possibilities
B 6 * 5 = 30
etc.
C 5 * 4 = 20
D 4 * 3 = 12
E 3 * 2 = 6
F 2 * 1 = 2
You stop at F because G and H have already been paired with all possible triplets.
total is sub of above 112
Aloha
2006-09-29 02:17:52 · answer #3 · answered by Anonymous · 1⤊ 1⤋
Urs is a permutation-combination problem.
Well u have to select 3 out of 8
therefore, 8C3 combinations will be available.
Now, 8C3 = 8!/(8-3)!*3! = 8!/5!*3! = 1*2*3*4*5*6*7*8 / (1*2*3*4*5)(1*2*3)
= 6*7*8 / 1*2*3
=7*8
= 56
This is ur answer.
Enjoy Maths.
2006-09-29 05:19:46 · answer #4 · answered by smilingbook1 2 · 1⤊ 0⤋
112.
to create any group of 3, the college has 8 students from which to choose the first assistant, 7 to choose the second, and 6 to choose the third.
that gives 8*7*6 = 336...
however, that would consider a group of students A, B, and C to be distinct from the groups of students B, C, and A, students A, C, and B, etc, when obviously any group of the same three students is in this case the same.
dividing by 3 gives the number of actually distinct groups, 112.
2006-09-29 04:45:09 · answer #5 · answered by Emily 3 · 0⤊ 1⤋
take 1 and 2 then you can put 6 students in the third seat.
take 1 and 3 then you can put 5 students
take 1 and 4 then you can out 4 students
1 and 5 3
1 6 2
1 7 1
now repeat for 2 54321
3 4321
4 321
5 21
6 1
so you get 2(6*1 + 5*2 + 4*3)=56 ways
2006-09-29 08:13:54 · answer #6 · answered by Anonymous · 0⤊ 0⤋
8C3 8!*3!/5! = (8/1)(7/2)(6/3) = 8*7
2006-09-29 02:17:14 · answer #7 · answered by Helmut 7 · 0⤊ 0⤋
3?
2006-09-29 02:16:58 · answer #8 · answered by mer_tice 1 · 0⤊ 3⤋
2.2
2006-09-29 02:18:03 · answer #9 · answered by Mer 2 · 0⤊ 1⤋