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Need help with this combination problem

by knight247 » Mon Jun 27, 2011 10:11 am
A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?

(A)45
(B)120
(C)80
(D)95
(E)30

OA is C


Its a very easy problem. Normal procedure is as follows:

'total number of possible communities' - 'total number of communities which have one married couple'

i.e.

10C3-(5C1*8C1)=80. But just hoping someone can help me solve without using the above method. I mean finding the total number of couple-less communities using a direct method. Even though it may be more complicated I would still like to clear out my basics. Detailed explanations would be appreciated. Thanks
Last edited by knight247 on Mon Jul 11, 2011 4:20 am, edited 1 time in total.
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by Frankenstein » Mon Jun 27, 2011 10:25 am
Hi,
There are 10 people
1st person can be picked in 10C1 = 10 ways
2nd person is to be picked from remaining 8(excluding the pair of 1st person) in 8C1 = 8 ways
3rd person is to be picked from remaining 6(excluding the pairs of 1st and 2nd persons) in 6C1 = 6 ways
Let the three picked be a,b,c
Now we are counting the same combination of a,b,c in different order as
a,b,c
c,b,a
b,c,a... so on in 3! ways.
so, number of communities that can be formed is 10*8*6/3! = 80

Hence, C
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by Ashley@VeritasPrep » Mon Jun 27, 2011 10:27 am
Here's what I'd do to get there directly: For the first person you choose, you have 10 possibilities. For the next person, you'll then be choosing from 8 possibilities (since you can't choose the person you already choose OR that person's spouse). For the next pick, you'll have six remaining possibilities to choose from (since another couple will have been taking out of the drawing pool). So from there we get 10*8*6. At that point I just need to account for the fact that these are combinations we want, not permutations -- so rather than letting any given group of three people count 3! times in its 3! different possible arrangements, I'll divide through by that 3! so I'll only be counting each group once. So (10*8*6)/3! -- and there you get to 80.
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by amit2k9 » Tue Jun 28, 2011 3:49 am
first person = 10
2nd person = 8
3rd person = 6 ways

10*8*6/ 3! (not an ordered pair ) = 80
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by kevincanspain » Tue Jun 28, 2011 7:03 am
Also

Step 1 : choose 3 of 5 couples 5c3 = 10

Step 2: for each of the 3 couples, choose either male or female : 2 x 2 x 2 = 8


Answer 10 x 8 = 80
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by Buix0065 » Tue Jun 28, 2011 7:23 pm
knight247 wrote:A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?

(A)45
(B)120
(C)80
(D)95
(E)30

OA is C


Its a very easy problem. Normal procedure is as follows:

'total number of possible communities' - 'total number of communities which have once married couple'

i.e.

10C3-(5C1*8C1)=80. But just hoping someone can help me solve without using the above method. I mean finding the total number of couple-less communities using a direct method. Even though it may be more complicated I would still like to clear out my basics. Detailed explanations would be appreciated. Thanks
Hi Knight,

In your original solve: 10C3-(5C1*8C1)

I'm actually lost on where 8C1 comes from, can you explain the (5C1*8C1) and how you arrive at that?

thank you!
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by Frankenstein » Tue Jun 28, 2011 8:34 pm
Buix0065 wrote: In your original solve: 10C3-(5C1*8C1)
I'm actually lost on where 8C1 comes from, can you explain the (5C1*8C1) and how you arrive at that?

thank you!
Hi,
5C1 accounts for 1 couple selected out of the 5 couples. So, out of the 3 positions to be filled, 2 are filled with the couple. We are left with 1 position to be filled with any of the other 8. It can be done in 8C1 ways.
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by Ashley@VeritasPrep » Wed Jun 29, 2011 9:34 am
Replying to a PM re: what the division by 3! is doing here.

We need to divide by 3! here because this problem is asking about combinations, not permutations. Let's say we start off just filling three slots from the original 10 people (5 couples).

Then I'd have 10 choices for who to put in my first slot. After I put someone there, I'd have 8 choices for who to put in my next slot (since whoever I initially chose AND that person's spouse would no longer be eligible). Then I'd have 6 remaining choices for who to put in my third, final slot. So, so far we're at 10*8*6 possibilities. Now, IF I were actually assigning specific ROLES -- e.g. if I were picking one person to be the leader of the committee, one person to be the secretary, and one person to be the alternate -- I'd stop there, and say there were 480 (that is, 10*8*6) possibilities for how I could assign these roles to three people, choosing from among my initial 10 with the no-more-than-one-from-each-couple restriction.

But that's not what I'm doing in this problem -- this problem just wants me to form a committee -- there are no specific roles. So now the thing is, say we look at one of our 10*8*6 possibilities, and it goes Alfred, Doris, Edward. Well, somewhere in that list of 10*8*6 possibilities, I'm also going to find a possible committee consisting of Alfred, Edward, Doris. And I'm also somewhere going to find a possible committee consisting of Doris, Alfred, Edward. And one consisting of Doris, Edward, Alfred. Then I'll find one consisting of Edward, Alfred, Doris... and finally one consisting of Edward, Doris, Alfred. And the catch is that these really shouldn't be counted as 6 different possible committees, because they're all actually the same. Since for any set of three people I chose, I would have found that set 6 times on my list, I divide through by 6 to get rid of all those extra copies.

Specifically, it wound up being 6 here that I needed to divide through by because 6 -- 3! -- is the number of possible arrangements of any given group of three people. If I were choosing 4-person committees instead, I'd instead divide whatever initial product I'd gotten by 4!, because my initial list would have had any given group of four people showing up in 4! different arrangements. In short, any given group of n particular people can show up in n! different arrangements, so when you're dealing with combinations (rather than permutations) problems, in which all of those n! arrangements amount to the same exact group, you divide through by n! so that you only count that group once instead of n! times.

Let me know if that clears it up!
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