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If a committee of 3 people is to be selected from among 5 [#permalink]
05 Jan 2010, 06:46

00:00

A

B

C

D

E

Difficulty:

25% (low)

Question Stats:

67% (01:47) correct
32% (01:13) wrong based on 168 sessions

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?

Re: committee of 3 [#permalink]
05 Jan 2010, 07:06

1

This post received KUDOS

ANS -80.. total people=10.. ways to select 3 out of them=10c3=120... it includes comb including couple.. ways in which couple are included =8c1*5=40.. so ans reqd 120-40=80... (if we take a gp to include a couple ,it will include couple +any one of rest 8 so 8c1 ways .. 5 couple so 5*8c1=40)

Re: committee of 3 [#permalink]
05 Jan 2010, 08:03

3

This post received KUDOS

kirankp 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? A. 20 B. 40 C. 50 D. 80 E. 120

total no for selecting 3 out of 10=10c3=120

no. of ways in which no two married people included= tot- 2 married couple included 2 married couple can be included in 5c1( no. of ways selecting a couple) * 8c1( no. of ways selecting the third person)=5 * 8=40

reqd comb=120-40=80 hence D
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Re: committee of 3 [#permalink]
05 Jan 2010, 09:07

10

This post received KUDOS

Expert's post

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?

Another way to think about this problem:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Re: committee of 3 [#permalink]
06 Jan 2010, 10:34

5

This post received KUDOS

I like to think of it like this:

Step 1 - find the combinations without any restrictions

10C3 = 120

Step 2 - subtract the combinations that would have a couple in the committee

5C1 x 4C1 x 2 = 40

In this step, we first find the # of ways to choose a couple, which is 5C1=5. After getting the first couple, we need 1 more member, so we choose 1 couple of the remainin 4 couples, which is 4C1 = 4. But within this new couple, we can either choose the man or the woman, so we need to x2.

Step 3 - find answer (no restrictions minus restrictions)

Re: committee of 3 [#permalink]
06 Jan 2010, 10:37

Bunuel 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?

Another way to think about this problem:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Re: committee of 3 [#permalink]
28 Sep 2010, 07:38

Bunuel 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?

Another way to think about this problem:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

I like this way of thinking and the calculations seem simpler and quicker.

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
21 Aug 2013, 16:36

Can someone please help me? I don't know what I am doing wrong.

5C1*2C1*4C1*2C1*8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple 4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple 8C1 - people that can occupy the third spot

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
22 Aug 2013, 02:23

Expert's post

brunawang wrote:

Can someone please help me? I don't know what I am doing wrong.

5C1*2C1*4C1*2C1*8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple 4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple 8C1 - people that can occupy the third spot

Re: committee of 3 [#permalink]
26 Aug 2013, 08:38

mrblack wrote:

I like to think of it like this:

Step 1 - find the combinations without any restrictions

10C3 = 120

Step 2 - subtract the combinations that would have a couple in the committee

5C1 x 4C1 x 2 = 40

In this step, we first find the # of ways to choose a couple, which is 5C1=5. After getting the first couple, we need 1 more member, so we choose 1 couple of the remainin 4 couples, which is 4C1 = 4. But within this new couple, we can either choose the man or the woman, so we need to x2.

Step 3 - find answer (no restrictions minus restrictions)

120 - 40 = 80

So the answer is 80.

i don't understand. We need one more member, so we choose 1 “couple"? (hence 2 members)

Re: committee of 3 [#permalink]
27 Aug 2013, 03:12

Expert's post

oss198 wrote:

mrblack wrote:

I like to think of it like this:

Step 1 - find the combinations without any restrictions

10C3 = 120

Step 2 - subtract the combinations that would have a couple in the committee

5C1 x 4C1 x 2 = 40

In this step, we first find the # of ways to choose a couple, which is 5C1=5. After getting the first couple, we need 1 more member, so we choose 1 couple of the remainin 4 couples, which is 4C1 = 4. But within this new couple, we can either choose the man or the woman, so we need to x2.

Step 3 - find answer (no restrictions minus restrictions)

120 - 40 = 80

So the answer is 80.

i don't understand. We need one more member, so we choose 1 “couple"? (hence 2 members)

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
25 Oct 2013, 18:01

Using slot method: First person can be chosen -> 10 ways, 2nd person can be chosen -> 8 ways (1st person and his wife are not candidates) and 3rd person can be chosen -> 6 ways (1st person/2nd person and their wives are out) Answer -> 10*8*6/6 = 80 (divide by 6 because the 3 people can be chosen in any order (i.e. 3! = 3*2*1 ways))