pairing combination

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

pairing combination

by shashank.ism » Tue Feb 09, 2010 1:06 pm

There are 10 students out of which three are boys and seven are girls. In how many different ways can the students be paired such that no pair consists of two boys?

A) 630
B) 1260
C) 105
D) 210
E) None

Last edited by shashank.ism on Wed Feb 10, 2010 9:44 am, edited 1 time in total.

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Mom4MBA: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Fri Jan 22, 2010 1:10 pm; Location: New Jersey; Thanked: 13 times; Followed by:4 members; GMAT Score:640

by Mom4MBA » Tue Feb 09, 2010 2:45 pm

I don't know how to do it, I have done it in two ways and getting different answers, can any one help with the right solution and explanation.

first method:
there are 3 boys and 7 girls, we have to make pairs with boys not together.

first boy goes with any of the 7 girls
second boy can go with any of the 6 girls
third boy can go with any of the 5 girls

now there are 4 girls left

so out of these one can go with any of the 3 left

at last there are 2 left to be paired

in all we have 7x6x5x3= 630

Second method:

10C2 - 3C2 = 42

I am more inclined towards this

Stay focused

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Wed Feb 10, 2010 9:57 am

Mom4MBA wrote:I don't know how to do it, I have done it in two ways and getting different answers, can any one help with the right solution and explanation.

first method:
there are 3 boys and 7 girls, we have to make pairs with boys not together.

first boy goes with any of the 7 girls
second boy can go with any of the 6 girls
third boy can go with any of the 5 girls

now there are 4 girls left

so out of these one can go with any of the 3 left

at last there are 2 left to be paired

in all we have 7x6x5x3= 630

Second method:

10C2 - 3C2 = 42

I am more inclined towards this

To make a valid pair a boy must be paired with a girl. First
choose a girl partner for each of the three boys.
The number of ways choosing 3 girls is = 7C3
The number of ways the chosen three girls can be paired
with three boys is = 3! Ã— 7C3
The number of ways in which the remaining four girls can be
paired among themselves
4C2/2!
Hence, total number of ways= (3! Ã— 7C3) Ã— 3= 630

So Mom4MBA you were correct in ur 1st approach .

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Mom4MBA: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Fri Jan 22, 2010 1:10 pm; Location: New Jersey; Thanked: 13 times; Followed by:4 members; GMAT Score:640

by Mom4MBA » Wed Feb 10, 2010 4:12 pm

Thanks Shashank

Stay focused

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tienvunguyen: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Thu Jul 30, 2009 6:58 am; Location: Madison, WI; Thanked: 1 times; Followed by:2 members; GMAT Score:750

by tienvunguyen » Thu Feb 18, 2010 8:31 am

There is a piece in this problem I don't understand. How can you have 630 pairs when the maximum of pairs taken out of 10 people is 10C2=45?

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tienvunguyen: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Thu Jul 30, 2009 6:58 am; Location: Madison, WI; Thanked: 1 times; Followed by:2 members; GMAT Score:750

by tienvunguyen » Thu Feb 18, 2010 8:33 am

There is a piece in this problem I don't understand. How can you have 630 pairs when the maximum of pairs taken out of 10 people is 10C2=45?

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

you left out the other 4 x 2 pairing

by harsh.champ » Thu Feb 18, 2010 9:13 am

tienvunguyen wrote:There is a piece in this problem I don't understand. How can you have 630 pairs when the maximum of pairs taken out of 10 people is 10C2=45?

I guess your post just got duplicated over here.

As for the question,yes 10C2 =45 but what about the rest of the 8 people.They can also be paired up differently.

So u will get like 10C2 x 8C2 x 6C2 x 4C2 x 2C2 but over here I have also considered the case when 2 boys are paired up together.

I guess you left out the pairing of the other 8 ppl and hence got only 10C2.
Hope this helps you to understand the question carefully.

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