gmat1011 wrote:

There is a category of GMAT questions where combinations related to 'pair/couples' is tested. This may also apply in a probability framework. Can people please add to this thread any tricky Pairs, Twins and Couples problems they spot during the course of their study. I want to have them all in one place as a reference resource and also to revise in the future before G-Day. Thanks!

EXAMPLE 1:

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?

Ans is 80.

METHOD 1: 10 * 8 * 6 / 3! is one way to the answer. If you don't divide by the factorial of the number of spots filled up you will be over counting.

METHOD 2: 10C3 - 5C1 * 8C1 (total number of ways to select 3 people out of 10 minus the product of ways to select 1 couple and one other person from the rest)

EXAMPLE 2: Example from MGMAT:

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

8/33

62/165

17/33

103/165

25/33

Method 1:

17/33 by doing 1 - P(no similar pairs) = 1 - 12/12 * 10/11 * 8/10 * 6/9 = 1 - 480/990 = 1-16/33 = 17/33

Method 2:

12 * 10 * 8 * 6/ 4! should be the numerator of the probability (desired events) = 240 = number of ways to select 4 cards which are all distinct

Total ways to select 4 cards out of 12 = 12C4 = 495

Prob = 1 - 240/495 = 1 - 16/33 = 17/33

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

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

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

Answer: D.

Similar problems with different approaches:

combination-permutation-problem-couples-98533.html?hilit=married%20couplesps-combinations-94068.html?hilit=married%20couplescommittee-of-88772.html?hilit=married%20couples2. Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?A. 8/33

B. 62/165

C. 17/33

D. 103/165

E. 25/33[/quote]

Let's calculate the opposite probability ans subtract this value from 1.

Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \(\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\).

\(C^4_6\) - # of ways to choose 4 different cards out of 6 different values;

\(2^4\) - as each of 4 cards chosen can be of 2 different suits;

\(C^4_{12}\) - total # of ways to choose 4 cards out of 12.

So \(P=1-\frac{16}{33}=\frac{17}{33}\).

Or another way:We can choose any card for the first one - \(\frac{12}{12}\);

Next card can be any card but 1 of the value we'v already chosen - \(\frac{10}{11}\) (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left);

Next card can be any card but 2 of the values we'v already chosen - \(\frac{8}{10}\) (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left);

Last card can be any card but 3 of the value we'v already chosen - \(\frac{6}{9}\);

\(P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\).

So \(P=1-\frac{16}{33}=\frac{17}{33}\) - the same answer as above.

Answer: C.

Hope it helps.

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