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GMAT Prep Combo Question - pls answer specific question

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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GMAT Prep Combo Question - pls answer specific question

by san2009 » Thu Aug 05, 2010 12:59 pm
This question has been solved but I have a specific question that has NOT been addressed. My exam is next week - will appreciate expedited response.

A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

When I looked at this question, I wanted to apply the SLOT method. And this is how I thought about it
3 slots: we must have one senior partner, and the other 2 slots can be filled by any body
4*9*8=288
Now, since order does NOT matter -- we're picking "groups" of 3 here...we must divide by the factorial of the number of inter-changeable items.
There are 3 inter-changeable items.
Let senior partner = SP and junior partner = JP
SP, JP, JP
JP, SP, JP
JP, JP, SP
*are all the same
Thus, we divide 288/3! = 48
Note: my interpretation of a group is that it doesn't matter how the 3 items are arranged within that group. And it doesn't matter what those 3 items are.
The OA is 100.

HOW and WHY is my approach incorrect?
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by Stuart@KaplanGMAT » Thu Aug 05, 2010 1:19 pm
san2009 wrote:This question has been solved but I have a specific question that has NOT been addressed. My exam is next week - will appreciate expedited response.

A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

When I looked at this question, I wanted to apply the SLOT method. And this is how I thought about it
3 slots: we must have one senior partner, and the other 2 slots can be filled by any body
4*9*8=288
Now, since order does NOT matter -- we're picking "groups" of 3 here...we must divide by the factorial of the number of inter-changeable items.
There are 3 inter-changeable items.
Let senior partner = SP and junior partner = JP
SP, JP, JP
JP, SP, JP
JP, JP, SP
*are all the same
Thus, we divide 288/3! = 48
Note: my interpretation of a group is that it doesn't matter how the 3 items are arranged within that group. And it doesn't matter what those 3 items are.
The OA is 100.

HOW and WHY is my approach incorrect?
The problem is that not every case will have the same number of duplicate arrangements. For this type of question, it's essential to treat each case separately.

Using the slot method:

1 Sr + 2 Jrs: 4*6*5, but we have 2 duplicates, so we must divide by 2!... 120/2 = 60 possible groups;
2 Srs + 1 Jr: 4*3*6, but we have 2 duplicates, so we must divide by 2!... 72/2 = 36 possible groups; and
3 Srs: 4*3*2, but we have 3 duplicates, so we must divide by 3!... 24/6 = 4 possible groups.

60 + 36 + 4 = 100 possible groups in total.

Of course, we could also have used combinations for each scenario:

1 Sr + 2 Jrs: 4C1 * 6C2 = 4*15 = 60;
2 Srs + 1 Jr: 4C2 * 6C1 = 6*6 = 36; and
3 Srs: 4C3 = 4.

60 + 36 + 4 = 100 possible groups in total.
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by san2009 » Thu Aug 05, 2010 1:51 pm
Stuart Kovinsky wrote:
san2009 wrote:This question has been solved but I have a specific question that has NOT been addressed. My exam is next week - will appreciate expedited response.

A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

When I looked at this question, I wanted to apply the SLOT method. And this is how I thought about it
3 slots: we must have one senior partner, and the other 2 slots can be filled by any body
4*9*8=288
Now, since order does NOT matter -- we're picking "groups" of 3 here...we must divide by the factorial of the number of inter-changeable items.
There are 3 inter-changeable items.
Let senior partner = SP and junior partner = JP
SP, JP, JP
JP, SP, JP
JP, JP, SP
*are all the same
Thus, we divide 288/3! = 48
Note: my interpretation of a group is that it doesn't matter how the 3 items are arranged within that group. And it doesn't matter what those 3 items are.
The OA is 100.

HOW and WHY is my approach incorrect?
The problem is that not every case will have the same number of duplicate arrangements. For this type of question, it's essential to treat each case separately.

Using the slot method:

1 Sr + 2 Jrs: 4*6*5, but we have 2 duplicates, so we must divide by 2!... 120/2 = 60 possible groups;
2 Srs + 1 Jr: 4*3*6, but we have 2 duplicates, so we must divide by 2!... 72/2 = 36 possible groups; and
3 Srs: 4*3*2, but we have 3 duplicates, so we must divide by 3!... 24/6 = 4 possible groups.

60 + 36 + 4 = 100 possible groups in total.

Of course, we could also have used combinations for each scenario:

1 Sr + 2 Jrs: 4C1 * 6C2 = 4*15 = 60;
2 Srs + 1 Jr: 4C2 * 6C1 = 6*6 = 36; and
3 Srs: 4C3 = 4.

60 + 36 + 4 = 100 possible groups in total.

Can you pls elaborate on:
"The problem is that not every case will have the same number of duplicate arrangements. For this type of question, it's essential to treat each case separately."

How and why is that so?
No matter what you do, you will be picking 3 individuals. Those 3 individuals can be arranged within any order WITHIN the group. If the individuals are SP, JP, JP or JP, SP, SP --> it shouldn't matter. In each scenario you will have 3 people who can be arranged in 3 different ways. thus you divide by 3!

Thanks a bunch!!
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by Stuart@KaplanGMAT » Thu Aug 05, 2010 5:48 pm
san2009 wrote: Can you pls elaborate on:
"The problem is that not every case will have the same number of duplicate arrangements. For this type of question, it's essential to treat each case separately."

How and why is that so?
No matter what you do, you will be picking 3 individuals. Those 3 individuals can be arranged within any order WITHIN the group. If the individuals are SP, JP, JP or JP, SP, SP --> it shouldn't matter. In each scenario you will have 3 people who can be arranged in 3 different ways. thus you divide by 3!

Thanks a bunch!!
No - when you have SP JP JP, for example, you have 2 duplicate entities - the JPs. When you have SP SP SP, you have 3 duplicate entities, the SPs. By breaking it down into different cases you account for the different duplications.

Let's call the SPs A, B, C and D and the JPs E, F, G, H, I and J.

Using your proposed method, one possible arrangement would be:

A E F

However, using your slot method,

E F A

would NOT be a possible selection, since by assigning the first slot a value of "4" you've already stipulated that only A, B, C and D are eligible for that slot.

So, when you divide through by 3!, you're factoring out possibilities that you've already eliminated.
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by san2009 » Fri Aug 06, 2010 2:35 am
hmm...that's a different approach then what i've learned -- confused, i am

are we not supposed to divide by the number of inter-changeable items?

it seems that you are saying, we should divide by the duplicates only, correct?
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by Stuart@KaplanGMAT » Fri Aug 06, 2010 8:08 am
san2009 wrote:hmm...that's a different approach then what i've learned -- confused, i am

are we not supposed to divide by the number of inter-changeable items?

it seems that you are saying, we should divide by the duplicates only, correct?
The problem is that you seem to be mixing approaches.

Under your initial approach, you calculated:

4 * 9 * 8

recognizing that there were only 4 people that could to in the first slot. In other words, not all slots were equal. Since not all slots were equal, you can't simply pretend that they are and divide by 3!.

If the question had simply been:

3 people are to be chosen out of 10, how many subgroups can be formed?

Then you could have thought:

OK - I have 10 available for the first slot, 9 for the second slot and 8 for the 3rd slot, so that's 10*9*8. However, since order doesn't matter, I need to eliminate all of the duplicate arrangements. Since there are 3 interchangeable slots, I have to divide by 3!. Then you'd correctly get:

10 * 9 * 8 / 3!

In fact, the calculation above is simply a derivation of the combinations formula:

10C3 = 10!/7!3! = 10*9*8*7!/7!3! = 10*9*8/3!

So again, here's the simplest rule to follow: when you're solving for alternative cases, count the number of options for each case and then add them together.
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by Jill » Fri Aug 13, 2010 1:57 am
Hi,

Realize this response is coming a bit too late, but would just like to confirm if my approach to the problem is correct also:-

Q: How many groups of 3 with at least 1 senior partner as a member?

This means that the only NOT possible answer would be if all members were only junior partners:-
i.e. 6C3 = 6!/(3!(3!)) = 20

Total possible choices of groups of 3 (not taking into account senior or junior members):-
10C3 = 10!/(3!(7!)) = 120

Therefore, answer would be (total possible choices - not possible choice) = (120 - 20) = 100 (b).

Thanks!
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