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by anirudhbhalotia » Thu Nov 25, 2010 6:54 am
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I got this question in a practice, was completely lost as to whats the answer, but more importantly whats the concept behind it.

Thanks for the help!

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by rishab1988 » Thu Nov 25, 2010 8:43 am
Question - > 4 person task force.

Total no of men->4
No of men to be chosen for task force->2

No. of ways of choosing 2 men [order does not matter]-> 4C2 -> 6

Total no of women ->3
No of women to be chosen for task force->2

No. of ways of choosing 2 women ->3C1-> 3

Total no of ways of choosing 2 men and 2 women -> 6*3 ->18[because of the word and ,if its or then we use +]

Hence answer ->B

Alternatively,assume M denotes men and W denotes women.

Then the no of ways of MMWW->(4*3*3*2)/2*2-> 18

2 men-> 4*3 [ we have 4 different ways to choose first man and 3 different to choose from the second man].Divide by 2 beacause order does not matter.

Similarly ,2 women ->3*2 and then divide by 2

Hence B

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by anirudhbhalotia » Thu Nov 25, 2010 10:42 pm
rishab1988 wrote:Question - > 4 person task force.

Total no of men->4
No of men to be chosen for task force->2

No. of ways of choosing 2 men [order does not matter]-> 4C2 -> 6

Total no of women ->3
No of women to be chosen for task force->2

No. of ways of choosing 2 women ->3C1-> 3

Total no of ways of choosing 2 men and 2 women -> 6*3 ->18[because of the word and ,if its or then we use +]

Hence answer ->B

Alternatively,assume M denotes men and W denotes women.

Then the no of ways of MMWW->(4*3*3*2)/2*2-> 18

2 men-> 4*3 [ we have 4 different ways to choose first man and 3 different to choose from the second man].Divide by 2 beacause order does not matter.

Similarly ,2 women ->3*2 and then divide by 2

Hence B


This is a bouncer for me. Is it a Permutation/Combination problem?

How do u get 4C2->6?

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by rishab1988 » Fri Nov 26, 2010 2:39 am
This can never be a permuation question for the follwoing reason

1) You are choosing a group. If you teacher asked you to name the students[[A,B,C] in your group ,then writing first A then B and then C is not different from writing B A C because the people in the group wont change.
2) Use permutations only when the question conveys an idea of ordering .Eg positions in a race.Ranking of countries in terms of GOLD medals in Asian games,a three digit number and so on.
3) Use Combinations only when orderings dont matter.For eg in a restaurant ordering chicken and then rice is not different from ordering rice and chicken [ you are ordering the same 2 items].Do you ever say to the waiter "I want rice and chicken but not chicken and rice?" Sounds absurd right. Similarly you use combinations when you have to pick a team [ lets assume you have to pick 11 players for your team.Then if you change the order of their names in a press release, will your team composition change?".Same scenario whn choosing a committe.

In general,think of this as,does 1st 2nd 3rd matter? If not COMBINATION if it does then PERMUTATION.

Using the same concept in your example.

Lets assume you have to choose two men from 4 in the group and you use the concept of permutation.

Let M1 and M2 denote the two men.

Then,according to permutations, M1 M2[two men] is different from M2 M1 [two men].So ask yourself,am I really changing the group by changing their order.NO.Because you are choosing the same people.

If you used the concept of Combinations,you assume that the 2 men are inherently the same.Therefore,M1M2 = M2M1.We therefore divide the no of permutations by 2 for repeats.The Permutation formula is 4P2 -> 12 ways. Now divide by 2 for repeats.No of ways = 6.This is same as the combination formula 4C2-> 6.