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P & C

by prachi18oct » Sun Jul 12, 2015 9:53 am
Tom is arranging his marble collection in a collector's case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must at least have one of each type, how many different ways can he arrange the case?
A 120
B 150
C 420
D 1,260
E 1,680
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by MartyMurray » Sun Jul 12, 2015 8:26 pm
There are two things going on here. One is the various combinations of marbles. The other is the arrangement of the combinations chosen.

Let's call the three types of marbles C, S and A.

The first thing we notice is that any combination is going to include at least 1 C, 1 S, and 1 A. So there are just two slots up for grabs.

So each combination will start this way. C S A _ _

To fill those two slots we have 4 C, 4 S, and 2 A to from which to choose.

We could figure out how many unique ways there are to fill those slots using a formula, but we need to take the additional step of arranging the marbles in each group. So we need to know the composition of each group. So I am just going to write out the possible groups.

C S A C C
C S A C S
C S A C A
C S A S S
C S A S A
C S A A A

That's all of them. It didn't really matter that we had 4 C and 4 S to choose from as there are only two slots to fill.

Now to find the number of possible arrangements of each of the six combinations.

C S A C C 5!/3! = 20
C S A C S 5!/2!2! = 30
C S A C A 5!/2!2! = 30
C S A S S 5!/3! = 20
C S A S A 5!/2!2! = 30
C S A A A 5!/3! = 20

20 + 30 + 30 + 20 + 30 + 20 = 150

Choose B.

Probably this BellCurves question takes slightly more work than you will have to do to answer any question on the actual GMAT, but this question is definitely good practice, because among other things in order to answer it you need to see the ramifications of having at least one of each, you need to see that you need to know the composition of each group, and you need to know how to calculate the number of possible permutations of each group.
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by prachi18oct » Mon Jul 13, 2015 7:06 am
Marty Murray wrote:There are two things going on here. One is the various combinations of marbles. The other is the arrangement of the combinations chosen.

Let's call the three types of marbles C, S and A.

The first thing we notice is that any combination is going to include at least 1 C, 1 S, and 1 A. So there are just two slots up for grabs.

So each combination will start this way. C S A _ _

To fill those two slots we have 4 C, 4 S, and 2 A to from which to choose.

We could figure out how many unique ways there are to fill those slots using a formula, but we need to take the additional step of arranging the marbles in each group. So we need to know the composition of each group. So I am just going to write out the possible groups.

C S A C C
C S A C S
C S A C A
C S A S S
C S A S A
C S A A A

That's all of them. It didn't really matter that we had 4 C and 4 S to choose from as there are only two slots to fill.

Now to find the number of possible arrangements of each of the six combinations.

C S A C C 5!/3! = 20
C S A C S 5!/2!2! = 30
C S A C A 5!/2!2! = 30
C S A S S 5!/3! = 20
C S A S A 5!/2!2! = 30
C S A A A 5!/3! = 20

20 + 30 + 30 + 20 + 30 + 20 = 150

Choose B.

Probably this BellCurves question takes slightly more work than you will have to do to answer any question on the actual GMAT, but this question is definitely good practice, because among other things in order to answer it you need to see the ramifications of having at least one of each, you need to see that you need to know the composition of each group, and you need to know how to calculate the number of possible permutations of each group.
Why didn't we first choose the marbles from the given set as in if the arrangement is C S A C C then for filling the marbles I have 5C1 * 5C1 * 3C1 * 4C1 * 3C1 and then we arrange them in 5!/3! ways . SO shouldn't it be like that? Why is the selcetion not important here?

Please explain.
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by GMATGuruNY » Mon Jul 13, 2015 7:46 am
prachi18oct wrote:Tom is arranging his marble collection in a collector's case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must at least have one of each type, how many different ways can he arrange the case?
A 120
B 150
C 420
D 1,260
E 1,680
When an arrangement includes IDENTICAL ELEMENTS, we must divide by the number of ways each set of identical elements can be arranged.

Let X, Y and Z represent the 3 marble types in the arrangement.
Since AT LEAST ONE of each marble type must be selected, two cases are possible.

Case 1: XXXYZ (3 of one marble type, 1 of each of the other two types)
Here, we must be divide by 3! to account for the 3 identical X's.
One marble type must be chosen to serve as X: the marble type that appears 3 times.
Number of options for marble X = 3. (Any of the 3 marble types.)
Number of ways to arrange XXXYZ = 5!/3! = 20.
To combine these options, we multiply:
3*20 = 60.

Case 2: XXYYZ (2 of one marble type, 2 of another marble type, 1 of the remaining type)
Here, we must be divide by 2! to account for the 2 identical X's and by another 2! to account for the 2 identical Y's.
2 marble types must be chosen to serve as X and Y: the 2 marble types that each appear twice.
Number of ways to choose 2 marble types from 3 options = 3C2 = (3*2)/(2*1) = 3.
Number of ways to arrange XXYYZ = 5!/(2!2!) = 30.
To combine these options, we multiply:
3*30 = 90.

Total ways = Case 1 + Case 2 = 60+90 = 150.

The correct answer is B.
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by MartyMurray » Mon Jul 13, 2015 8:03 am
prachi18oct wrote: Why didn't we first choose the marbles from the given set as in if the arrangement is C S A C C then for filling the marbles I have 5C1 * 5C1 * 3C1 * 4C1 * 3C1 and then we arrange them in 5!/3! ways . SO shouldn't it be like that? Why is the selcetion not important here?

Please explain.
This is not a probability question the answering of which requires that we determine by how many different paths we can get to a set or an arrangement. Answering this question requires only figuring out how many possible different sets there are and then arranging them.

Look at this. What if we have 6 marbles, three of them red and three blue? All of the red and all of the blue may look the same as each other, but they are not the same marble. If we want to get two red and we are drawing two marbles, we could pick the first two red ones, the second two red ones, the first and last red ones, the second then the first red one, and so on. So even though each time we are getting two red marbles, there are actually multiple paths via which that can happen.

In the question we are discussing here, however, we don't care how many paths there are to a certain arrangement. We only care about how many unique arrangements there are. So we don't care that, for instance, there are five paths to putting C in the first slot.

Further, we may have five C's, but at most we can use 3 C's in one arrangement. So, those other two C's don't matter because we don't care that we have them. We only care that we can make arrangements that include one, two, or three C's.

Overall, the question is not about how many ways can we arrange all of the 15 marbles we have. It's about how many different arrangements we can make given the marbles we have and the constraints. In fact the answer to this question would be exactly the same if we had 50 of each type of marble. We still would get 150 possible unique arrangements given the constraints.

We start with C S A, and then we can fill the other slots with one each of two types or two of one type. That's it. So, as long as we have at least three of each type, the answer to the question will be unaffected by the number of marbles we have to work with.
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by Sun Light » Wed Jul 15, 2015 2:27 am
prachi18oct wrote:Tom is arranging his marble collection in a collector's case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must at least have one of each type, how many different ways can he arrange the case?
A 120
B 150
C 420
D 1,260
E 1,680
Murray cracked it.

I was also initially confused with the approach. I think the key is the word 'different ways' which tells us to apply permutation.

The following example might help in further understanding of this logic:

Company X has 6 regional offices. Each regional office must recommend two candidates, one male and one female, to serve on the corporate auditing committee. If each of the offices must be represented by exactly one member on the auditing committee and if the committee must consist of an equal number of male and female employees, how many different committees can be formed?

Ans is 20

i will try to add more questions related to this particular topic..

hope it helps.
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by prachi18oct » Wed Jul 15, 2015 6:34 am
Sun Light wrote:
prachi18oct wrote:Tom is arranging his marble collection in a collector's case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must at least have one of each type, how many different ways can he arrange the case?
A 120
B 150
C 420
D 1,260
E 1,680
Murray cracked it.

I was also initially confused with the approach. I think the key is the word 'different ways' which tells us to apply permutation.

The following example might help in further understanding of this logic:

Company X has 6 regional offices. Each regional office must recommend two candidates, one male and one female, to serve on the corporate auditing committee. If each of the offices must be represented by exactly one member on the auditing committee and if the committee must consist of an equal number of male and female employees, how many different committees can be formed?

Ans is 20

i will try to add more questions related to this particular topic..

hope it helps.
Can uou explain how did you arrive at given OA for your question?
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by Sun Light » Wed Jul 15, 2015 9:47 am
prachi18oct wrote:
Sun Light wrote:
prachi18oct wrote:Tom is arranging his marble collection in a collector's case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must at least have one of each type, how many different ways can he arrange the case?
A 120
B 150
C 420
D 1,260
E 1,680
Murray cracked it.

I was also initially confused with the approach. I think the key is the word 'different ways' which tells us to apply permutation.

The following example might help in further understanding of this logic:

Company X has 6 regional offices. Each regional office must recommend two candidates, one male and one female, to serve on the corporate auditing committee. If each of the offices must be represented by exactly one member on the auditing committee and if the committee must consist of an equal number of male and female employees, how many different committees can be formed?

Ans is 20

i will try to add more questions related to this particular topic..

hope it helps.
Can uou explain how did you arrive at given OA for your question?
1."How many different committees can be formed" --> Permutation.

2."If each of the offices must be represented by exactly one member on the auditing committee and if the committee must consist of an equal number of male and female employees. Note: 6 offices = 6 members.

Therefore, 6 members can be arranged in 6! ways. 3 female and 3 male must be there in the group.

So, Final answer:- 6!/(3!*3!) = 20.

"Selection" of people from offices doesn't play a role in this approach.

Six office - Six people.. out of 6, 3 are female and 3 are males...



i will share more problems on arrangement in this post....
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by vineet.nitd » Wed Jul 15, 2015 9:25 pm
In such problems, one should try to satisfy the condition first. The condition here is that at least one marble of each type should be in the case.

So, we give one marble of each type to the case.

Now we are left with suit that can accommodate two more marbles out of the remaining marbles: 4Cs, 4Ss, 2As

The number of cases of allocation from the three types will be whole number solutions of the equation a+b+c = 2 and they are 4C2 or 6 in number.(011, 020, 002, 101, 110, 200)

We can have any of the above 6 cases going into the suit and then once we have that case in the suit, the arrangement needs to be done.

3 * {5!/(2!*2!) + 5!/(3!)} = 3*50 = 150 [first term is for cases, in which one marble each of two types go into the case, and second term is for cases with both marbles of one type going into the case]
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by Amrabdelnaby » Wed Nov 18, 2015 11:28 am
Hi Guru,

Can you please explain your approach here a little further.

I solved it using listing and i got 150 but it took more steps and time than your method.

Could you please explain a bit further to me your method.

Much appreciated :)
GMATGuruNY wrote:
prachi18oct wrote:Tom is arranging his marble collection in a collector's case. He has five identical cat-eyes, five identical sulphides, and three identical agates. If he can fit exactly five marbles into the case and must at least have one of each type, how many different ways can he arrange the case?
A 120
B 150
C 420
D 1,260
E 1,680
When an arrangement includes IDENTICAL ELEMENTS, we must divide by the number of ways each set of identical elements can be arranged.

Let X, Y and Z represent the 3 marble types in the arrangement.
Since AT LEAST ONE of each marble type must be selected, two cases are possible.

Case 1: XXXYZ (3 of one marble type, 1 of each of the other two types)
Here, we must be divide by 3! to account for the 3 identical X's.
One marble type must be chosen to serve as X: the marble type that appears 3 times.
Number of options for marble X = 3. (Any of the 3 marble types.)
Number of ways to arrange XXXYZ = 5!/3! = 20.
To combine these options, we multiply:
3*20 = 60.

Case 2: XXYYZ (2 of one marble type, 2 of another marble type, 1 of the remaining type)
Here, we must be divide by 2! to account for the 2 identical X's and by another 2! to account for the 2 identical Y's.
2 marble types must be chosen to serve as X and Y: the 2 marble types that each appear twice.
Number of ways to choose 2 marble types from 3 options = 3C2 = (3*2)/(2*1) = 3.
Number of ways to arrange XXYYZ = 5!/(2!2!) = 30.
To combine these options, we multiply:
3*30 = 90.

Total ways = Case 1 + Case 2 = 60+90 = 150.

The correct answer is B.
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