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Combinatorics- 600-700 level

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Combinatorics- 600-700 level

by lavinia » Fri Oct 01, 2010 6:03 am
How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
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by kmittal82 » Fri Oct 01, 2010 6:32 am
>How many different 5-person teams can be formed from a group of x individuals?

= x!/[5!*(x-5)!]. If we find x, we can calculate this

>(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(x+2)!/[5!*(x-3)!] = 126

(x+2)(x+1)x(x-1)(x-2)(x-3)!
---------------------------------- = 126
5! * (x-3)!

=> (x+2)(x+1)x(x-1)(x-2) = 126 x 5!

This equation will have more than 1 root, so insufficient.

>(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

(x+1)!/[3!(x-2)!] = 56

(x+1)x(x-1)(x-2)!
--------------------- = 56
3! * (x-2)!

x(x+1)(x-1) = 56x3x2 = 336

x = 7 is the only solution for this, so (B) is sufficient
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by kmittal82 » Fri Oct 01, 2010 6:40 am
I just rechecked using a calculator, and x= 7 is also a solution for equation 1, but how can we possibly know that by just looking at the equation?
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by lavinia » Fri Oct 01, 2010 10:16 am
The answer choice is D I'm still waiting for an explanation. Thanks kmittal82.
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by GMATGuruNY » Fri Oct 01, 2010 11:42 am
lavinia wrote:How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
Total number of possible combinations = (the product of descending consecutive integers)/(number of elements being chosen)!

For example, the number of 4-member groups that can be made from 6 individuals = 6C4 = 6*5*4*3/4! = 15.

The number of integers in the numerator will be the same as the number of elements being chosen. The total number of individuals will be the largest of the descending consecutive integers in the numerator.

In the scenario above, we have 4 elements in our combination, so there are 4 descending consecutive integers in the numerator. The largest of the consecutive integers (6) = the total number of individuals.

Now let's tackle the DS question above. Let P = the product of the descending consecutive integers in the numerator.

Statement 1:
P/5! = 126
P = 126 * 5! = (9*7*2)*(5*4*3*2*1)
P = 9*8*7*6*5
Thus, if we have 9 people, we can form (9*8*7*6*5)/5! = 126 groups.
So x+2=9, and x=7.
Sufficient.

Statement 2:
P/3! = 56
P = (8*7)*(3*2*1)
P = 8*7*6
Thus, if we have 8 people, we can from 8*7*6/3! = 56 groups.
So x+1=8, and x=7.
Sufficient.

The correct answer is D.
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by lavinia » Sun Oct 03, 2010 12:47 pm
Wow it's a great explanation. Now it makes sense.

Thank you GmatGuruNY!
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by lavinia » Mon Oct 04, 2010 9:32 am
I still have a question regarding this problem. Did you apply this formula: N!/ K!(N-K)!?


GMATGuruNY wrote:
lavinia wrote:How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
Total number of possible combinations = (the product of descending consecutive integers)/(number of elements being chosen)!

For example, the number of 4-member groups that can be made from 6 individuals = 6C4 = 6*5*4*3/4! = 15.

The number of integers in the numerator will be the same as the number of elements being chosen. The total number of individuals will be the largest of the descending consecutive integers in the numerator.

In the scenario above, we have 4 elements in our combination, so there are 4 descending consecutive integers in the numerator. The largest of the consecutive integers (6) = the total number of individuals.

Now let's tackle the DS question above. Let P = the product of the descending consecutive integers in the numerator.

Statement 1:
P/5! = 126
P = 126 * 5! = (9*7*2)*(5*4*3*2*1)
P = 9*8*7*6*5
Thus, if we have 9 people, we can form (9*8*7*6*5)/5! = 126 groups.
So x+2=9, and x=7.
Sufficient.

Statement 2:
P/3! = 56
P = (8*7)*(3*2*1)
P = 8*7*6
Thus, if we have 8 people, we can from 8*7*6/3! = 56 groups.
So x+1=8, and x=7.
Sufficient.

The correct answer is D.
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by GMATGuruNY » Mon Oct 04, 2010 11:11 am
lavinia wrote:I still have a question regarding this problem. Did you apply this formula: N!/ K!(N-K)!?


GMATGuruNY wrote:
lavinia wrote:How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
Total number of possible combinations = (the product of descending consecutive integers)/(number of elements being chosen)!

For example, the number of 4-member groups that can be made from 6 individuals = 6C4 = 6*5*4*3/4! = 15.

The number of integers in the numerator will be the same as the number of elements being chosen. The total number of individuals will be the largest of the descending consecutive integers in the numerator.

In the scenario above, we have 4 elements in our combination, so there are 4 descending consecutive integers in the numerator. The largest of the consecutive integers (6) = the total number of individuals.

Now let's tackle the DS question above. Let P = the product of the descending consecutive integers in the numerator.

Statement 1:
P/5! = 126
P = 126 * 5! = (9*7*2)*(5*4*3*2*1)
P = 9*8*7*6*5
Thus, if we have 9 people, we can form (9*8*7*6*5)/5! = 126 groups.
So x+2=9, and x=7.
Sufficient.

Statement 2:
P/3! = 56
P = (8*7)*(3*2*1)
P = 8*7*6
Thus, if we have 8 people, we can from 8*7*6/3! = 56 groups.
So x+1=8, and x=7.
Sufficient.

The correct answer is D.
I had the formula in mind but found it easier to solve as I did above. Applying the combinations formula to statement 1, we'd get:

N!/(5!(N-5)!)= 126
N!/(N-5)! = 126 * 5! = 9*7*2*5*4*3*2*1 = 9*8*7*6*5
9!/4! = 9*8*7*6*5
N=9
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by lavinia » Tue Oct 05, 2010 10:52 am
GMATGuruNY wrote:
lavinia wrote:I still have a question regarding this problem. Did you apply this formula: N!/ K!(N-K)!?


GMATGuruNY wrote:
lavinia wrote:How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
Total number of possible combinations = (the product of descending consecutive integers)/(number of elements being chosen)!

For example, the number of 4-member groups that can be made from 6 individuals = 6C4 = 6*5*4*3/4! = 15.

The number of integers in the numerator will be the same as the number of elements being chosen. The total number of individuals will be the largest of the descending consecutive integers in the numerator.

In the scenario above, we have 4 elements in our combination, so there are 4 descending consecutive integers in the numerator. The largest of the consecutive integers (6) = the total number of individuals.

Now let's tackle the DS question above. Let P = the product of the descending consecutive integers in the numerator.

Statement 1:
P/5! = 126
P = 126 * 5! = (9*7*2)*(5*4*3*2*1)
P = 9*8*7*6*5
Thus, if we have 9 people, we can form (9*8*7*6*5)/5! = 126 groups.
So x+2=9, and x=7.
Sufficient.

Statement 2:
P/3! = 56
P = (8*7)*(3*2*1)
P = 8*7*6
Thus, if we have 8 people, we can from 8*7*6/3! = 56 groups.
So x+1=8, and x=7.
Sufficient.

The correct answer is D.
I had the formula in mind but found it easier to solve as I did above. Applying the combinations formula to statement 1, we'd get:

N!/(5!(N-5)!)= 126
N!/(N-5)! = 126 * 5! = 9*7*2*5*4*3*2*1 = 9*8*7*6*5
9!/4! = 9*8*7*6*5
N=9
Thank you!
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