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Combination Doubt

by gmatmillenium » Tue Jun 15, 2010 12:37 am
A 3-people committee is to be selected from four couples. If the committee cannot contain a couple at the same time, how many such committees are possible?

The explanation to this (I believe by one of the experts) was given as below

in the first slot you can pick any of the 8 people - so 8 possibilities.

in the second slot there are only 6 people - you already chose one and you can't pick the one from the same couple.

For the last slot there are 4 people available - from the last two couples.

So 8*6*4/(no. of slots !) - for a total of 32

My Query is - why the division by (no. of slots !)??
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by kvcpk » Tue Jun 15, 2010 1:10 am
Imagine 4 couples as 4 boxes with each having m,w

we need to choose 3 persons(either mor w) from 4 boxs.

So we are trying to choose 3 boxes in particular wich can be done in 4c3 ways.
And Since u have 2 choices per couple & there are 3 couples, u get
4c3*2*2*2
=32

Hope this helps!!
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by gmatmillenium » Tue Jun 15, 2010 1:43 am
if you're assuming one couple as one entity and doing a 4c3, you are selection 6 people, not 3

thanks for trying but I hvnt followed this




kvcpk wrote:Imagine 4 couples as 4 boxes with each having m,w

we need to choose 3 persons(either mor w) from 4 boxs.

So we are trying to choose 3 boxes in particular wich can be done in 4c3 ways.
And Since u have 2 choices per couple & there are 3 couples, u get
4c3*2*2*2
=32

Hope this helps!!
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by Testluv » Tue Jun 15, 2010 1:56 am
My Query is - why the division by (no. of slots
We divide by the number of possible arrangements because ABC is the same trio of people as BAC. That is, we divide by the number of arrangements because order doesn't matter. We want the number of unordered arrangements. To go from numbered of ordered arrangements to the number of unordered arrangements, we have to divide the number of arrangements by all the possible ways of arranging.

In the solution you referenced, we are "fixing" certain people in slots: We have 8 possible people in the first slot. Let's say we put Andy in the first slot. In the second slot, you can't have Andy or his wife, Charlotte. So, you can't have AC in the first two slots. But we could have Andy in the first slot, and Beth, who is not his wife, in the second slot. So, we can have AB in the first two slots. But couldn't we also just have Beth in the first slot, and Andy in the second? Can't we have BA in the first two slots? But are "AB" and "BA" different selections of people? Clearly not. That is, order doesn't matter.

As discussed, we can have any 8 in the first slot. Once we select one of these 8 for the first slot, we can't have that person or his/her significant other in the second slot. That leaves us with 6 for the second slot. As for the third slot, we can't have either of the 2 we picked for the first 2 slots; nor can we have their significant others. That leaves us with 4 for the third slot. So we have 8*6*4 selections. However, this solution must be incomplete. It assumes order is important and, therefore, only gives us the number of ORDERED selections.

But the order of this trio doesn't matter. So, we have to divide by 3! (ie, the number of ways of ordering/arranging).
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by kvcpk » Tue Jun 15, 2010 2:06 am
gmatmillenium wrote:if you're assuming one couple as one entity and doing a 4c3, you are selection 6 people, not 3

thanks for trying but I hvnt followed this
Hi,

Selecting one person is same as selecting his/her spouse too. So infact each time we are selecting a couple. As you say we are actaully choosing 6 people. to make sure that none of them get chosen again. But of them only 3 people will be considered.

Let me put it in more clear terms:
A,WA,B,WB,C,WC,D,WD be the pairs.

Suppose in the first chance, A is chosen. then WA will return to home. she is no longer part of the game.
Similarly when B is chosen, WB will return to home
When C, WC is out.

So, for choosing any 3 of the existing 4 pairs, we need 4c3 ways.

And I could choose either A or WA in each round. So multiply by 2c1 =2
hence we arrive at 4c3*2c1*2c1*2c1 = 32.

Hope this helps!!
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by GMATGuruNY » Tue Jun 15, 2010 2:12 am
When given a combination question in which some of the combinations will be unacceptable, remember:

good + bad = total

To determine the number of good options, sometimes the easiest approach is to figure out the total number of options and subtract the bad ones.

In this case:

Total = total number of 3-member committees that can be made from the 8 people without regard to the 4 couples.

Bad = the number of committees that combine one of the couples with a third person.

Total - Bad = Good

Total: The total number of 3-member committees that can be made from 8 people is 8 * 7 * 6 / 1 * 2 * 3 = 56 total committees.

Bad: Combining one of the 4 couples with one of the 6 remaining people gives us 4 * 6 = 24 bad committees. (It's as though we have 4 shirts and 6 ties and we're determining the number of outfits that can be made: 4 * 6 = 24).
.
So subtracting the 24 bad commmittees from the 56 total committees, we're left with 56 - 24 = 32 good committees.
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by gmatmillenium » Tue Jun 15, 2010 3:46 am
Thanks much Testluv....if I go by your dictum, then I obviously would not need to divide if lets say the question asked for how many ways can I arrange a 6 color strip on a flag fm 4 colors if say two colors cannot be together...am I right?


Testluv wrote:
My Query is - why the division by (no. of slots
We divide by the number of possible arrangements because ABC is the same trio of people as BAC. That is, we divide by the number of arrangements because order doesn't matter. We want the number of unordered arrangements. To go from numbered of ordered arrangements to the number of unordered arrangements, we have to divide the number of arrangements by all the possible ways of arranging.

In the solution you referenced, we are "fixing" certain people in slots: We have 8 possible people in the first slot. Let's say we put Andy in the first slot. In the second slot, you can't have Andy or his wife, Charlotte. So, you can't have AC in the first two slots. But we could have Andy in the first slot, and Beth, who is not his wife, in the second slot. So, we can have AB in the first two slots. But couldn't we also just have Beth in the first slot, and Andy in the second? Can't we have BA in the first two slots? But are "AB" and "BA" different selections of people? Clearly not. That is, order doesn't matter.

As discussed, we can have any 8 in the first slot. Once we select one of these 8 for the first slot, we can't have that person or his/her significant other in the second slot. That leaves us with 6 for the second slot. As for the third slot, we can't have either of the 2 we picked for the first 2 slots; nor can we have their significant others. That leaves us with 4 for the third slot. So we have 8*6*4 selections. However, this solution must be incomplete. It assumes order is important and, therefore, only gives us the number of ORDERED selections.

But the order of this trio doesn't matter. So, we have to divide by 3! (ie, the number of ways of ordering/arranging).
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by Testluv » Tue Jun 15, 2010 3:59 am
Thanks much Testluv....if I go by your dictum, then I obviously would not need to divide if lets say the question asked for how many ways can I arrange a 6 color strip on a flag fm 4 colors if say two colors cannot be together...am I right?
Correct. I've emboldened the keyword that tells us to consider order.

Incidentally, it may be helpful to you to notice that the only difference beween the nCk and nPk formulae is that the former also divides by n!.
Last edited by Testluv on Tue Jun 15, 2010 4:02 am, edited 1 time in total.
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by gmatmillenium » Tue Jun 15, 2010 4:01 am
bingo...thanks a lot testluv
Testluv wrote:
Thanks much Testluv....if I go by your dictum, then I obviously would not need to divide if lets say the question asked for how many ways can I arrange a 6 color strip on a flag fm 4 colors if say two colors cannot be together...am I right?
Correct. I've emboldened the keyword that tells us to consider order.
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by gmatmillenium » Tue Jun 15, 2010 4:37 am
Thanks Mitch....I used to employ the same logic...wonder why I didn't do the same again....
GMATGuruNY wrote:When given a combination question in which some of the combinations will be unacceptable, remember:

good + bad = total

To determine the number of good options, sometimes the easiest approach is to figure out the total number of options and subtract the bad ones.

In this case:

Total = total number of 3-member committees that can be made from the 8 people without regard to the 4 couples.

Bad = the number of committees that combine one of the couples with a third person.

Total - Bad = Good

Total: The total number of 3-member committees that can be made from 8 people is 8 * 7 * 6 / 1 * 2 * 3 = 56 total committees.

Bad: Combining one of the 4 couples with one of the 6 remaining people gives us 4 * 6 = 24 bad committees. (It's as though we have 4 shirts and 6 ties and we're determining the number of outfits that can be made: 4 * 6 = 24).
.
So subtracting the 24 bad commmittees from the 56 total committees, we're left with 56 - 24 = 32 good committees.
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by gmatmillenium » Tue Jun 15, 2010 5:10 am
Testluv,

May I request your reply to my other post too?....pasted here below

X range is -1 <X <1, Q Which of the following must be true?
Answers are:
1. X ^ 2> X ^ 3,
2. (X +1 / 2) (X-1 / 2) <0

Explanation given by one the experts was as under..

Obviously x^2 > x^3 does not hold for x = 0.
Taking the second statement we have that (X +1 / 2) (X-1 / 2) <0
So we have that either (a) (X +1 / 2)>0 and (X-1 / 2)< 0
Or
(b) (X +1 / 2)<0 and (X-1 / 2)>0. This would mean x<-1/2 and x>1/2 which is not possible and so (b) is automatically rejected.
So we have that (X +1 / 2) (X-1 / 2) <0 implies -1/2<x<1/2 which is possible because given range of x is -1 <x<1.

The second statement is true.

My query - for x= .6, statement 2 doesn't hold good so how is the answer b?

Testluv wrote:
My Query is - why the division by (no. of slots
We divide by the number of possible arrangements because ABC is the same trio of people as BAC. That is, we divide by the number of arrangements because order doesn't matter. We want the number of unordered arrangements. To go from numbered of ordered arrangements to the number of unordered arrangements, we have to divide the number of arrangements by all the possible ways of arranging.

In the solution you referenced, we are "fixing" certain people in slots: We have 8 possible people in the first slot. Let's say we put Andy in the first slot. In the second slot, you can't have Andy or his wife, Charlotte. So, you can't have AC in the first two slots. But we could have Andy in the first slot, and Beth, who is not his wife, in the second slot. So, we can have AB in the first two slots. But couldn't we also just have Beth in the first slot, and Andy in the second? Can't we have BA in the first two slots? But are "AB" and "BA" different selections of people? Clearly not. That is, order doesn't matter.

As discussed, we can have any 8 in the first slot. Once we select one of these 8 for the first slot, we can't have that person or his/her significant other in the second slot. That leaves us with 6 for the second slot. As for the third slot, we can't have either of the 2 we picked for the first 2 slots; nor can we have their significant others. That leaves us with 4 for the third slot. So we have 8*6*4 selections. However, this solution must be incomplete. It assumes order is important and, therefore, only gives us the number of ORDERED selections.

But the order of this trio doesn't matter. So, we have to divide by 3! (ie, the number of ways of ordering/arranging).
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by GMATGuruNY » Tue Jun 15, 2010 6:53 am
gmatmillenium wrote:Thanks Mitch....I used to employ the same logic...wonder why I didn't do the same again....
GMATGuruNY wrote:When given a combination question in which some of the combinations will be unacceptable, remember:

good + bad = total

To determine the number of good options, sometimes the easiest approach is to figure out the total number of options and subtract the bad ones.

In this case:

Total = total number of 3-member committees that can be made from the 8 people without regard to the 4 couples.

Bad = the number of committees that combine one of the couples with a third person.

Total - Bad = Good

Total: The total number of 3-member committees that can be made from 8 people is 8 * 7 * 6 / 1 * 2 * 3 = 56 total committees.

Bad: Combining one of the 4 couples with one of the 6 remaining people gives us 4 * 6 = 24 bad committees. (It's as though we have 4 shirts and 6 ties and we're determining the number of outfits that can be made: 4 * 6 = 24).
.
So subtracting the 24 bad commmittees from the 56 total committees, we're left with 56 - 24 = 32 good committees.
If we use the logic above, one advantage is that many times the number of BAD combinations will be small.

For example, if the problem above had stated that only one of the couples couldn't serve together, out of the TOTAL 56 committees we'd have only 1 * 6 = 6 BAD committees, leaving us with 56 - 6 = 50 GOOD committees.

Since the number of GOOD combinations often will be just a bit less than the TOTAL number of combinations, all we'll need to do is to determine the TOTAL and look for an answer choice that is just a bit smaller.
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by gmatmillenium » Tue Jun 15, 2010 7:19 am
Thanks Mitch
GMATGuruNY wrote:
gmatmillenium wrote:Thanks Mitch....I used to employ the same logic...wonder why I didn't do the same again....
GMATGuruNY wrote:When given a combination question in which some of the combinations will be unacceptable, remember:

good + bad = total

To determine the number of good options, sometimes the easiest approach is to figure out the total number of options and subtract the bad ones.

In this case:

Total = total number of 3-member committees that can be made from the 8 people without regard to the 4 couples.

Bad = the number of committees that combine one of the couples with a third person.

Total - Bad = Good

Total: The total number of 3-member committees that can be made from 8 people is 8 * 7 * 6 / 1 * 2 * 3 = 56 total committees.

Bad: Combining one of the 4 couples with one of the 6 remaining people gives us 4 * 6 = 24 bad committees. (It's as though we have 4 shirts and 6 ties and we're determining the number of outfits that can be made: 4 * 6 = 24).
.
So subtracting the 24 bad commmittees from the 56 total committees, we're left with 56 - 24 = 32 good committees.
If we use the logic above, one advantage is that many times the number of BAD combinations will be small.

For example, if the problem above had stated that only one of the couples couldn't serve together, out of the TOTAL 56 committees we'd have only 1 * 6 = 6 BAD committees, leaving us with 56 - 6 = 50 GOOD committees.

Since the number of GOOD combinations often will be just a bit less than the TOTAL number of combinations, all we'll need to do is to determine the TOTAL and look for an answer choice that is just a bit smaller.
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by gmatmillenium » Tue Jun 15, 2010 7:28 am
Mitch,

I will really appreciate if you can respond also to my today's post on "every prime number" - does gmat mean only distinct prime #s or all prime #s regardless of repitition?

gmatmillenium wrote:Thanks Mitch
GMATGuruNY wrote:
gmatmillenium wrote:Thanks Mitch....I used to employ the same logic...wonder why I didn't do the same again....
GMATGuruNY wrote:When given a combination question in which some of the combinations will be unacceptable, remember:

good + bad = total

To determine the number of good options, sometimes the easiest approach is to figure out the total number of options and subtract the bad ones.

In this case:

Total = total number of 3-member committees that can be made from the 8 people without regard to the 4 couples.

Bad = the number of committees that combine one of the couples with a third person.

Total - Bad = Good

Total: The total number of 3-member committees that can be made from 8 people is 8 * 7 * 6 / 1 * 2 * 3 = 56 total committees.

Bad: Combining one of the 4 couples with one of the 6 remaining people gives us 4 * 6 = 24 bad committees. (It's as though we have 4 shirts and 6 ties and we're determining the number of outfits that can be made: 4 * 6 = 24).
.
So subtracting the 24 bad commmittees from the 56 total committees, we're left with 56 - 24 = 32 good committees.
If we use the logic above, one advantage is that many times the number of BAD combinations will be small.

For example, if the problem above had stated that only one of the couples couldn't serve together, out of the TOTAL 56 committees we'd have only 1 * 6 = 6 BAD committees, leaving us with 56 - 6 = 50 GOOD committees.

Since the number of GOOD combinations often will be just a bit less than the TOTAL number of combinations, all we'll need to do is to determine the TOTAL and look for an answer choice that is just a bit smaller.
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