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Counting....Challenging
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knight247
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A company is selecting individuals to fill four seats on its board. Two of the seats will have major voting rights, and two will have minor voting rights. There is no difference between the two seats with major rights, nor is there a difference between the two seats with minor rights. The company's options are: from Division X: Alice and Bob; from Division Y: Carl and Diana; from Division Z: Edgar. Two people from the same division cannot get a board seat with the same type of voting rights. How many sets of board seat assignments are possible?
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GMATGuruNY
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So few assignments are possible that the easiest -- and quickest -- approach might be to write them out.knight247 wrote:A company is selecting individuals to fill four seats on its board. Two of the seats will have major voting rights, and two will have minor voting rights. There is no difference between the two seats with major rights, nor is there a difference between the two seats with minor rights. The company's options are: from Division X: Alice and Bob; from Division Y: Carl and Diana; from Division Z: Edgar. Two people from the same division cannot get a board seat with the same type of voting rights. How many sets of board seat assignments are possible?
In the groupings below, the first pair gets major voting rights, while the second pair gets minor voting rights.
The only pairings not allowed are AB and CD.
Without Edgar:
AC-BD
AD-BC
BC-AD
BD-AC.
Total options = 4.
With Edgar:
EA-BC
EA-BD
EB-AC
EB-AD
EC-AD
EC-BD
ED-AC
ED-BC
In each grouping the pairs can swap places, yielding 8 more options.
Total options = 16.
Total possible assignments = 4+16 = 20.
Last edited by GMATGuruNY on Mon Feb 20, 2012 3:26 pm, edited 4 times in total.
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Mike@Magoosh
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Hi. I'm happy to provide the solution for this. 
This indeed is a difficult counting problem, one more difficult that what one might expect to see on the real GMAT.
I'll consider four slot arrangements, with the first two having the major voting rights, and the last two having the minor voting rights.
The first two slots can have eight combinations, four of which involve E (the red ones):
AC, AD, AE, BC, BD, BE, CE, DE
If E is not chosen, then that means there's one from Division X and one from Division Y in the first pair. That leaves three people from three different divisions. Any combination would work, so there are 3C2 = 3 possibilities. For those four green pairs, there are 3 possible second pairs, for a total of 12.
If E is chosen in the first pair (e.g. AE), that means two of the remaining three people are from the same division and can't both be picked. Instead of three possible pairs from those remaining people, there would only be two --- for example, if AE is the first pair, then either BC or BD could be the second pair, but not CD. For the four red pairs, there are 2 possible second pairs, for a total of 8.
That gives 12+8=20 possibilities total. Since there are only 20, I'll list them (first pair has major voting rights, second pair has minor).
(AC BD), (AC BE), (AC DE), (AD BC), (AD BE),
(AD CE), (AE BC), (AE BD), (BC AD), (BC AE),
(BC DE), (BD AC), (BD AE), (BD AE), (BE AC),
(BE AD), (CE AD), (CE BD), (DE AC), (DE BC)
Does all that make sense? Please let me know if anyone reading this has any questions.
Mike
This indeed is a difficult counting problem, one more difficult that what one might expect to see on the real GMAT.
I'll consider four slot arrangements, with the first two having the major voting rights, and the last two having the minor voting rights.
The first two slots can have eight combinations, four of which involve E (the red ones):
AC, AD, AE, BC, BD, BE, CE, DE
If E is not chosen, then that means there's one from Division X and one from Division Y in the first pair. That leaves three people from three different divisions. Any combination would work, so there are 3C2 = 3 possibilities. For those four green pairs, there are 3 possible second pairs, for a total of 12.
If E is chosen in the first pair (e.g. AE), that means two of the remaining three people are from the same division and can't both be picked. Instead of three possible pairs from those remaining people, there would only be two --- for example, if AE is the first pair, then either BC or BD could be the second pair, but not CD. For the four red pairs, there are 2 possible second pairs, for a total of 8.
That gives 12+8=20 possibilities total. Since there are only 20, I'll list them (first pair has major voting rights, second pair has minor).
(AC BD), (AC BE), (AC DE), (AD BC), (AD BE),
(AD CE), (AE BC), (AE BD), (BC AD), (BC AE),
(BC DE), (BD AC), (BD AE), (BD AE), (BE AC),
(BE AD), (CE AD), (CE BD), (DE AC), (DE BC)
Does all that make sense? Please let me know if anyone reading this has any questions.
Mike
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GMATGuruNY
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If Edgar is NOT chosen:raunekk wrote:Can there be a shorter method for this?
Alice must be receive either major or minor voting rights.
Number of options for Alice = 2. (Major or minor.)
Number of people who could be paired with Alice = 2. (Carl or Diana).
Number of options for the remaining two positions = 1. (Must be taken by whoever is left.)
To combine these options, we multiply:
2*2*1 = 4.
If Edgar is chosen:
Number of options for Edgar = 2. (Major or minor).
Number of people who could be paired with Edgar = 4. (Any of the other 4 people).
The remaining two positions must include someone from the division not yet used.
Number of options from the division not yet used = 2.
Number of options for the last position = 1. (Of the 2 people left, we can't use the coworker of the third person chosen, leaving us only 1 option.)
To combine these options, we multiply:
2*4*2*1 = 16.
Total possible assignments = 4+16 = 20.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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Mike@Magoosh
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The bad news is: this is a difficult counting problem, and there's not really a shorter way to do this.raunekk wrote:Can there be a shorter method for this?
The good news is: if you think this is long, then you should see some really challenging problems that would take 10x longer than this little one. Be thankful you don't have to worry about those problems. This one, by comparison, is nothing.
The better news: even this problem is harder than what you will have to compute on the GMAT. The GMAT tends not to give problems that you have to solve by creating extensive lists. They tend to favor problems that lend themselves to more elegant solutions.
I hope that helps.
Mike
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kul512
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Lets say there are People X1, X2, Y1, Y2 and Z.
1st option-
in group 1- we take one person from X group, One from Y group. So total such option- 2C1X2C1
now one member of X and one member of Y group is left. Also Z member is left. So for forming 2nd group we can pick any of these three- 3C2
So total option-
2C1X2C1X3C2=12
2nd option-
In group one, one member is from either group X or group Y and Z is picked as second member. so total option- 2C1(Pick any one group from X and Y)X 2C1(pick a member from two members of the chosen group). Now for forming second group have only group X and Y and one group has two members left. we will pick one out of these two members so 2C1 and rest will be from other group.
2C1X2C1X2C1=8
Total=12+8=20
1st option-
in group 1- we take one person from X group, One from Y group. So total such option- 2C1X2C1
now one member of X and one member of Y group is left. Also Z member is left. So for forming 2nd group we can pick any of these three- 3C2
So total option-
2C1X2C1X3C2=12
2nd option-
In group one, one member is from either group X or group Y and Z is picked as second member. so total option- 2C1(Pick any one group from X and Y)X 2C1(pick a member from two members of the chosen group). Now for forming second group have only group X and Y and one group has two members left. we will pick one out of these two members so 2C1 and rest will be from other group.
2C1X2C1X2C1=8
Total=12+8=20
