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A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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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)

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

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.

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

Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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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]

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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.

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

Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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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

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]

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24 Jun 2016, 11:34

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Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

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24 Jun 2016, 13:02

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

Ways to choose at least 1 Senior

1 S and 2 J 4C1*6C2= 60 2 S and 1 J 4C2 * 6C1= 36 3 S and 0 J 4C3 = 4

60+36+4= 100

B is the answer
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Unfortunately, your math includes some 'duplicate entries.'

For example, let's call the 4 senior partners A, B, C and D and the 6 junior partners 1, 2, 3, 4, 5 and 6.

In your calculation, you state that the first person selected MUST be one of those 4 seniors (A/B/C/D) and the remaining two people can be any two of the remaining 9...

The group "A/B/1" and "B/A/1" are the SAME group, but your calculation counts THAT group TWICE (depending on whether A or B was chosen first). In a Combination question, you can't allow duplicate entries.

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