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A committe of three people is to be chosen from four married [#permalink]

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26 Jul 2008, 02:36

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A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

let me try...out of 4 couples we have to choose three at a time....no of combinations is given by 4C3 morever within each choosen couple we have a option of choosing one out of two people or 2C1

another method : calculate desired probablity and multiply it by total number of ways = 8C3 = 56

probablity of selecting first person = 8/8 (can be anyone) probability of selecting secd person = 6/7 ( there are 7 left, but we can not take spouse of first person) probability of selecting third person = 4/6 (there are 6 left but we can take only 4)

number of ways to choose first person : 8 number of ways to choose second person : 6 (cannot choose from the first couple) number of ways to choose third person : 4 (cannot choose from the first and 2nd couple)

therefore, number of ways = 8 * 6 * 4

however, this is the number of permutations (as {couple 1 husband, couple 2 husband, couple 3 husband} same as {couple 2 husband, couple 1 husband, couple 3 husband} )

therefore, divide by 3! therefore, number of ways = 8 * 6 * 4 / 3! = 32

A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16 2. 24 3. 26 4. 30 5. 32

Source -> GMAT prep

We have three positions to fill. Choice of first position = 8 (let's call him Mr. Bush) Choice of 2nd position = 6 (because we already chose Mr. Bush in first round and we cannot choose Mrs. Bush) , let call this 2nd committe member as Mr. Clinton. Choice of 3rd position = 4 (because we Mr. Bush and Mr. Clinton are already in the committee, and we cannot choose Mrs Bush or Mrs. Clinton)

A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16 2. 24 3. 26 4. 30 5. 32

Source -> GMAT prep

We have three positions to fill. Choice of first position = 8 (let's call him Mr. Bush) Choice of 2nd position = 6 (because we already chose Mr. Bush in first round and we cannot choose Mrs. Bush) , let call this 2nd committe member as Mr. Clinton. Choice of 3rd position = 4 (because we Mr. Bush and Mr. Clinton are already in the committee, and we cannot choose Mrs Bush or Mrs. Clinton)

Combination is 8x6x4 = 192

Not in answer choice.

what am I doing????

When u take 8x6x4, u r bringing arrangement (or order or position) into the expression. Since order of selection does not matter here, u need to divide 8x6x4 by the possible number of arrangements of 3 ppl. 3!=6. that will give u the true answer.

A committe of three people is to be chosen from four married couples . WHat is number of different committees that can be chosen if two people who are married to each other cannot both serve on committee

1. 16 2. 24 3. 26 4. 30 5. 32

Source -> GMAT prep

We have three positions to fill. Choice of first position = 8 (let's call him Mr. Bush) Choice of 2nd position = 6 (because we already chose Mr. Bush in first round and we cannot choose Mrs. Bush) , let call this 2nd committe member as Mr. Clinton. Choice of 3rd position = 4 (because we Mr. Bush and Mr. Clinton are already in the committee, and we cannot choose Mrs Bush or Mrs. Clinton)

Combination is 8x6x4 = 192

Not in answer choice.

what am I doing????

When u take 8x6x4, u r bringing arrangement (or order or position) into the expression. Since order of selection does not matter here, u need to divide 8x6x4 by the possible number of arrangements of 3 ppl. 3!=6. that will give u the true answer.

Thanks bhushangiri, but I think i only partly understand. Can you think of some examples?

Suppose u need to randomly select 2 out of 4 ppl.. A,B,C,D.

By your approach you would say, you have 4x3=12 ways.

But let us see... AB, AC, AD BC, BD, CD are the only possible ways. ----- (1)

When u consider 4x3, you bring in the possibilities of BA, CA, DA, CB, DB, and DC. When u have to simply select and not arrange, these are same as what is mentioned in (1).

a general point to remember.. when u are asked to "choose" or "select" then try and go with the combination approach i.e. using nCr formula. If you happen to use n(n-1)(n-2)... approach, then remember to divide by appropriate factorial term.

Last edited by bhushangiri on 28 Jul 2008, 11:34, edited 2 times in total.

Suppose u need to randomly select 2 out of 4 ppl.. A,B,C,D.

By your approach you would say, you have 4x3=12 ways.

But let us see... AB, AC, AD BC, BD, CD are the only possible ways. ----- (1)

When u consider 4x3, you bring in the possibilities of BA, CA, DA, CB, DB, and DC. When u have to simply select and not arrange, these are same as what is mentioned in (1).

a general point to remember.. when u are asked to "choose" or "select" then try and go with the combination approach i.e. using nCr formula. If you happen to use n(n-1)(n-2)... approach, then remember to divide by appropriate factorial term.