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# In a certain group of 10 members, 4 members teach only

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Manager
Joined: 17 Aug 2005
Posts: 167
Followers: 1

Kudos [?]: 58 [0], given: 0

In a certain group of 10 members, 4 members teach only [#permalink]  18 Oct 2005, 08:08
In a certain group of 10 members, 4 members teach only French and the rest teach only Spanish or German. If the group is to choose a 3-member committee, which must have at least 1 member who teaches French, how many different committees can be chosen?

(A) 40
(B) 50
(C) 64
(D) 80
(E) 100

VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 46 [0], given: 0

E. 100

Possible combinations are:
1F & 2 Others
2F & 1 Other
3F & 0 Other

total combinations:
4c1 * 6c2 + 4c2 * 6c1 + 4c3 * 6c0 = 100
Director
Joined: 14 Oct 2003
Posts: 588
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: Perms & Combs [#permalink]  18 Oct 2005, 08:36
kimmyg wrote:
In a certain group of 10 members, 4 members teach only French and the rest teach only Spanish or German. If the group is to choose a 3-member committee, which must have at least 1 member who teaches French, how many different committees can be chosen?

(A) 40
(B) 50
(C) 64
(D) 80
(E) 100

Easiest and shortest way to solve a combination with a constraint is to

First Find the Total Combination w/o constraints

10 C 3 = 120

Find the Combination of NO French teachers 6 C 3 = 20

Subtract the the Combination of having no French teachers from the total

10C3-6C3=100

Less than 1.5 mins.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5068
Location: Singapore
Followers: 23

Kudos [?]: 194 [0], given: 0

We're given:

French - 4 members
Spanish or German - 6 members

The group of three must have at least 1 member who teaches French:

Three cases:

Case 1: 2 French member and 1 Spanish or German

4C2*6C1 = 36 combinations

Case 2: 1 French member and 2 Spanish or German
4C1*6C2 = 60 combinations

Case 3: 3 French members
4C3 = 4 combinations

Total = 100 combinations
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