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A firm has 4 senior partners and 6 junior partners. How many

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A firm has 4 senior partners and 6 junior partners. How many [#permalink] New post 09 Dec 2010, 07:21
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A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)

A. 48
B. 100
C. 120
D. 288
E. 600
[Reveal] Spoiler: OA
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Re: Some PS problems [#permalink] New post 09 Dec 2010, 08:38
Expert's post
gdk800 wrote:
1) A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)
a. 48
b. 100
c. 120
d. 288
e. 600

[Reveal] Spoiler:
OA is B.


Total # of different groups of 3 out of 10 people: C^3_{10}=120;
# of groups with only junior partners (so with zero senior memeber): C^3_6=20;

So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100.

Answer: B.
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senior partners [#permalink] New post 20 Apr 2013, 10:22
What seems to work as well, though its tougher to come up with this solution:

(10*9*8) * 2/3 + (10*9*4)*1/3 = 600. Divided by all possible permutations (=3!) yields 100.

Explanation: For the first spot you have 10 possible canditates, for the 2nd 9. For the 3rd spot you need to differentiate if a senior partner has been picked for one of the first 2 spots or not. If yes, then you can pick one of the 8 remaining guys, so 10*9*8. If no senior partner has been picked yet, you need 10*9*4.
Now you need to weight the 2 cases with the corresponding probabilities: probability of no senior partner being picked in the first 2 draws = 6/10*5/9=1/3, so you weight 10*9*4 with 1/3. For the complementary case (senior partner was picked in the first 2 draws) you just take the complementary prob (1-1/3)= 2/3 and weight 10*9*8 with it.
Now you just need to divide the answer (600) by the number of different positions (=3!) and get 600/6=100

But I suggest you stick with the easy solution in the GMAT :)
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] New post 20 Apr 2013, 17:57
The standard approach

1 senior partner
4 x 6C2 = 60

2 senior partners
4C2 x 6 = 36

3 senior partners
4C3 = 4

Total 100
However GDKs approach is a better way around
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] New post 22 Apr 2013, 21:44
gdk800 wrote:
A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)

A. 48
B. 100
C. 120
D. 288
E. 600


A general approach to solving Combination problems

Steps:

1. There are two larger groups, the senior partners and the junior partners.
2. The first larger group i.e., the senior partners is 4 in number. So letn1 be 4. The second larger group, ie, the junior partners is 6 in number. So n2 is 6.
3. The smaller group that is selected from the larger group of senior partners may be 1, 2, or 3 in number. So r1 is 1 or 2 or 3 . Correspondingly the other smaller group i.e.,r2 selected from the junior partners is 2 or 1 or 0 in number.
4. For each value ofr1 and the correspondingr2, compute the number of combinations which are4C1 * 6C2, 4C2 * 6C1 and4C3 * 6C0 being 60, 36 and 4 ways respectively.
5. The total number of combinations is therefore 60+36+4 = 100
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Re: A firm has 4 senior partners and 6 junior partners. How many   [#permalink] 22 Apr 2013, 21:44
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