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Kevin must select 3 friends from different social groups to invite to

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Re: Kevin must select 3 friends from different social groups to invite to [#permalink]
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MonkeyDDes wrote:
mSKR wrote:
5 F in music club--atleast one

MMM-》 5C3--》= 10--> this combination can not be selected as they would be from all same club ( against informaiton given ina rgument)
MMR-》 5C2 + 4= 14
MRR-> 5+ 4*3/2= 11

how the correct answer is 80?

We divide the solution into cases, knowing that the answer is based on the calculation of combinations

Case 1: Invite 3 friends from the music club.
Combinations: 5C3 = 10

Case 2: Invite 2 music friends and 1 reading friend.
Combinations: 5C2 * 4C1 = 10 * 4 = 40

Case 3: Invite 1 musician friend and 2 reading friends
Combinations: 5C1 * 4C2 = 5 * 6 = 30

Total: 10 + 40 + 30 = 80

Kevin must select 3 friends from different social groups to invite to his party
i think MMM should not be applicable

btw, i got my mistake. it should be * not +

thanks1
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Kevin must select 3 friends from different social groups to invite to [#permalink]

Solution

Given
• Kevin has 5 friends in a music club and 4 friends in a reading club.

To find
• Ways of selecting 3 friends such that at least one must be from the music club.

Approach and Working out

Since at least one of the invited friends must be from the music club, thus, we can have the following cases.
Case 1:
• Exactly one friend belongs to the music club. In this case, the rest two must be from the reading club.
• Thus, combinations = $$(5C_1)(4C_2) = 5×6 = 30$$
Case 2:
• Exactly two friends belong to the music club. In this case, the remaining one must be from the reading club.
• Thus, combinations = $$(5C_2)(4C_1) = 10×4 = 40$$
Case 3:
• All three friends belong the music club.
• Thus, combinations = $$5C_3 = 10$$

Total number of combinations = 30 + 40 + 10 = 80.

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Re: Kevin must select 3 friends from different social groups to invite to [#permalink]
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MonkeyDDes wrote:
Kevin must select 3 friends from different social groups to invite to his party. He has 5 friends in a music club, and 4 friends in a reading club. At least one of the invited friends must be from the music club. How many different combinations of friends can Kevin invite to his party?

A.30
B.40
C.60
D.70
E.80

At least one of the friends invited from music club = Total friends invited - no friends invited from music club
So, total = $$9C_3$$ = 84
No friends invited from music club = $$4C_3$$ = 4
So, at least one = 84 - 4 = 80

IMO, (E)!
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Re: Kevin must select 3 friends from different social groups to invite to [#permalink]
1 from music + 2 from music + all from music=

5C1*4C2+5C2*4C1+5C3 = 30+40+10 = 80
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Kevin must select 3 friends from different social groups to invite to [#permalink]
Can anyone explain why can we not calculate strait 5*8C2? reason behind is that 5C1=5 since 1 must be invited from music. and no other condition. so 8 people left in total and we need to choose 2 of them. so 8C2=28. 5*28=140. I understand this is flawed but can't figure out what am I missing here?

Maybe Bunuel could help?

update:

I think I got it now. the question is not really about friends. so friends are not different, they are labaled as music guy or reading guy. The question asks about combinations of clubs, not people. Question could be clearer regarding this matter.

Posted from my mobile device
Kevin must select 3 friends from different social groups to invite to [#permalink]
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