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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
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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 let\(n1\) 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 of\(r1\) and the corresponding\(r2\), compute the number of combinations which are\(4C1 * 6C2\), \(4C2 * 6C1\) and\(4C3 * 6C0\) being 60, 36 and 4 ways respectively.
5. The total number of combinations is therefore 60+36+4 = 100
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A firm has 4 senior partners and 6 junior partners. How many [#permalink] New post 28 Dec 2014, 07:49
You can also think about it as a code.

4S - 6J

three different ways to group them:

S J J. 4(6)5/2! because we have two junior members
or
S S J. 4(3)6/2! because we have two senior members
or
S S S 4(3)2/3! because we have three senior members.

add up the results and you get 100.
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Last edited by gmat6nplus1 on 28 Dec 2014, 09:38, edited 1 time in total.
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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] New post 28 Dec 2014, 09:07
Expert's post
Hi All,

This is a rare layered Combinatorics question. To answer it, you can either figure out the Total of ALL possibilities and subject the ones that "don't fit" or you can perform 3 separate calculations to account for the ones that "do fit." Here is how you can approach the latter option:

We're given 4 senior partners and 6 junior partners. We're asked for the number of different groups of 3 (the clue that we'll need the Combination Formula) with one stipulation - there must be AT LEAST 1 senior partner. Here are the 3 calculations:

1) 3 senior partners = 4c3 = 4!/[3!1!] = 4 different groups

2) 2 seniors and 1 junior = (4c2)(6c1) = (4!/[2!2!])(6!/[1!5!) = (6)(6) = 36 different groups

3) 1 senior and 2 juniors = (4c1)(6c2) = (4!/[3!1!])(6!/[2!4!]) = (4)(15) = 60 different groups

4 + 36 + 60 = 100 different groups

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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] New post 10 May 2015, 01:40
Bunuel wrote:
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.


Princeton book is wonderful for explanation of coounting and combination. read it to understand of the math legendary
Re: A firm has 4 senior partners and 6 junior partners. How many   [#permalink] 10 May 2015, 01:40
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