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committee

by ch0719 » Wed Dec 24, 2008 2:35 pm
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

20
40
50
80
120


Answer is 80

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Re: committee

by logitech » Wed Dec 24, 2008 7:36 pm
ch0719 wrote:If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

20
40
50
80
120


Answer is 80

thanks
How many different ways can you select 3 people out of 10 ?

10C3=120

Now lets find the number of ways that we can have married couple in the group and substract this from 120.

Lets say first couple took the first 2 spots. Now we have 8 people to choose from for the last spot = 8 ways

Since we have 5 couples:

8x5 = 40

So the solution to the problem is:

120-40 = 80
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by ch0719 » Wed Dec 24, 2008 9:55 pm
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by ronniecoleman » Wed Dec 24, 2008 10:17 pm
My answer is 80

There are 5 couples.....so 5 ways to choose one...

For a grup of 3 choose one couple in 5 ways...
8 people left choose one people in 8 ways
Total ways= 40 ( in which we have one married couple)

Total ways of choosing 3 people out of 10 10c3 = 120
Ways in which we have no couples 120-40
= 80
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by vittalgmat » Sat Dec 27, 2008 12:00 am
Here is another way, analytically. I found this easier and can do it fast as well (under 1 min)

let the married couples be denoted as
Aa Bb Cc Dd Ee

Let us write the 3 spots as below
Spot1 Spot2 Spot3



Now Spot 1 can be occupied by any of the 10 ppl
so
Spot 1 = 10 ways -------------------(1)

Spot 2 can be occupied by 8 ppl
(coz Spot 1's spouse cannot be on the committee)
So Spot 2 = 8 ways -------------------(2)

Similarly for Spot 3 we have 6 ways -----(3).

So total Arrangements (permutations) = 10 *8*6. ----(4)

Since (4) is order dependendent, we need to divide it by
3! (coz the size of the committee is 3. Generally, divide by factorial of committee size)

10*8*6 / (3*2) = 80

HT Helps.

Note:
If permutation/arrangement is asked we dont have to do the last step of dividing by 3!.
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by cramya » Sat Mar 07, 2009 4:38 pm
Nice approach, Vittal.

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CR
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by apple100 » Tue May 05, 2009 5:03 pm
vittalgmat wrote:Here is another way, analytically. I found this easier and can do it fast as well (under 1 min)

let the married couples be denoted as
Aa Bb Cc Dd Ee

Let us write the 3 spots as below
Spot1 Spot2 Spot3



Now Spot 1 can be occupied by any of the 10 ppl
so
Spot 1 = 10 ways -------------------(1)

Spot 2 can be occupied by 8 ppl
(coz Spot 1's spouse cannot be on the committee)
So Spot 2 = 8 ways -------------------(2)

Similarly for Spot 3 we have 6 ways -----(3).

So total Arrangements (permutations) = 10 *8*6. ----(4)

Since (4) is order dependendent, we need to divide it by
3! (coz the size of the committee is 3. Generally, divide by factorial of committee size)

10*8*6 / (3*2) = 80

HT Helps.

Note:
If permutation/arrangement is asked we dont have to do the last step of dividing by 3!.
What is the rule in dividing this equation by 3! ?
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by lilu » Tue May 05, 2009 7:54 pm
apple100 wrote:
vittalgmat wrote:Here is another way, analytically. I found this easier and can do it fast as well (under 1 min)

let the married couples be denoted as
Aa Bb Cc Dd Ee

Let us write the 3 spots as below
Spot1 Spot2 Spot3



Now Spot 1 can be occupied by any of the 10 ppl
so
Spot 1 = 10 ways -------------------(1)

Spot 2 can be occupied by 8 ppl
(coz Spot 1's spouse cannot be on the committee)
So Spot 2 = 8 ways -------------------(2)

Similarly for Spot 3 we have 6 ways -----(3).

So total Arrangements (permutations) = 10 *8*6. ----(4)

Since (4) is order dependendent, we need to divide it by
3! (coz the size of the committee is 3. Generally, divide by factorial of committee size)

10*8*6 / (3*2) = 80

HT Helps.

Note:
If permutation/arrangement is asked we dont have to do the last step of dividing by 3!.
What is the rule in dividing this equation by 3! ?
Because with the method above you are counting ABC, ACB, BCA, CAB.... when you don't need to because it doesn't matter how people in the committee are arranged.
The more you look, the more you see.
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by sureshbala » Tue May 05, 2009 11:35 pm
The first person can be selected in 10 ways

Once we select the first person leaving out the person who is related to the first person selected we can select the second person from remaining 8 persons in 8 ways.

Similarly after selecting the second person, we can select the third person from the remaining 6 persons in 6 ways.

Hence the total number of possible selections = 10 x 8 x 6 = 480.

But since the order is not to be considered here (i.e selecting ABC is same as selecting BCA) = 480/3! = 80 ways
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