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Junior | Next Rank: 30 Posts
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by reza_moohebat » Sat Aug 28, 2010 3:46 am
How should we solve this question?

A badminton team consist of 7 students. The team will be chosen from a group of 8 boys and 5 girls. Find the number of team that can be formed such that each team consists of

i.4 boys
ii.Not more than 2 girl
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by sirisha.g » Sat Aug 28, 2010 3:51 am
1) 4 boys=> 4 boys and 3 girls= 8c4*5c3

2)not more than one girl=> you can have 0 or 1 girl=> 8c7+8c6*5c1
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by reza_moohebat » Sat Aug 28, 2010 4:22 am
sirisha.g wrote:1) 4 boys=> 4 boys and 3 girls= 8c4*5c3

2)not more than one girl=> you can have 0 or 1 girl=> 8c7+8c6*5c1
I have the answers of these questions. Your first answer was correct my friend, but I have 708 for ii section and it is not correspond with your second answer.

I have another question. When we know we have similar things, like 4 boys, why should we should use combination? It seems all the boys are similar and it does not change anything which one is selected. If we had for example john, David, Frank and etc, combination was meaningful, but I cannot see any differences between our candidates. Can you explain me please.
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by debmalya_dutta » Sat Aug 28, 2010 4:51 am
sirisha.g wrote:1) 4 boys=> 4 boys and 3 girls= 8c4*5c3

2)not more than 2 girls => you can have 0 or 1 girl=> 8c7+8c6*5c1
the ii question one says not more than 2 girls

So number of ways = 0 girls, 7 boys + 1 girl,6 boys + 2girls,5 boys
= 8C7+5*8C6+5C2*8C5
@Deb
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by sirisha.g » Sat Aug 28, 2010 4:51 am
hey i misread the ii. question. i read it as not more than one girl.

so there can be 0 or 1 or 2 girls=> 8c7+ 8c6*5c1+8c5*5c2 =708

regarding your question. boys are not similar because one individual is different from another one.
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by diebeatsthegmat » Sat Aug 28, 2010 11:25 am
reza_moohebat wrote:How should we solve this question?

A badminton team consist of 7 students. The team will be chosen from a group of 8 boys and 5 girls. Find the number of team that can be formed such that each team consists of

i.4 boys
ii.Not more than 2 girl
1. 4 boys =>3 girls
[8!/(4!*4!) ]*5!/(2!*3!)

2. not more than 2 girls
we have to 7 students in which 1. there are 2 girls and 5 boy 2. there are 1 girls and 6 boys and 3. there are 0 girls and 7 boys
and we did the same as above then add all of em together
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by narik11 » Wed Sep 01, 2010 3:59 am
sirisha.g wrote:hey i misread the ii. question. i read it as not more than one girl.

so there can be 0 or 1 or 2 girls=> 8c7+ 8c6*5c1+8c5*5c2 =708

regarding your question. boys are not similar because one individual is different from another one.

the question is regarding the number of teams that can be formed.. and not the number of ways the team can be formed. ??
please correct me if am wrong..[/quote]
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