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29 Mar 2006, 12:43
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I'm having a tough time figuring out this one too. Any takers?
Every night, the Biased News Network (BNN) features more guests from the Bow Tie Party than from the Ascot Party. If 11 members of each party are available to appear on BNN tonight, do more than 4 members of the Ascot party get to appear?
(1) At least 462 different groups of the available Bow Tie Party members could be chosen to appear on BNN tonight.
(2) At least 330 different groups of the available Ascot Party members could be chosen NOT to appear on BNN tonight.
[/b]
Every night, the Biased News Network (BNN) features more guests from the Bow Tie Party than from the Ascot Party. If 11 members of each party are available to appear on BNN tonight, do more than 4 members of the Ascot party get to appear?
(1) At least 462 different groups of the available Bow Tie Party members could be chosen to appear on BNN tonight.
(2) At least 330 different groups of the available Ascot Party members could be chosen NOT to appear on BNN tonight.
[/b]
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29 Mar 2006, 17:54
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My answer would be (E)
We have the total no. of members available 22 (11 from each), but not the desire total to appear in the show. I think that is required to answer this question.
We have the total no. of members available 22 (11 from each), but not the desire total to appear in the show. I think that is required to answer this question.
I would say the answer is D.
This is because:
- From (1) we get ---------------------------------------------
11 Choose n = 462 = 11!/(n!/(11-n)!)
or (n!/(11-n!)) = 11!/462 = 39916800/462 = 86400.
where n is the number of Bow Tie Party member selected
By trying different values for n we get the value of n==5, satisfying the above equation. Since we know that Bow Tie Party is more than Ascot, Ascot has to have less than 5 (not more than 4) members.
.. in pluging values I tried, n=4,5 to make a quick decision.
- From (2) we get ---------------------------------------------
(11 - n)! - (11 Choose n) = 330
where n is the number of Ascot party members selected.
Knowing GMAT/ETS, usually the results of (1) are aligned with the results of the (2). So in the above equation i started with n=4. And Voila LHS == RHS.
So n if 4. So we know that the number of Ascot Party members is not more than 4 .
Hence, I chose Answer D.
This is because:
- From (1) we get ---------------------------------------------
11 Choose n = 462 = 11!/(n!/(11-n)!)
or (n!/(11-n!)) = 11!/462 = 39916800/462 = 86400.
where n is the number of Bow Tie Party member selected
By trying different values for n we get the value of n==5, satisfying the above equation. Since we know that Bow Tie Party is more than Ascot, Ascot has to have less than 5 (not more than 4) members.
.. in pluging values I tried, n=4,5 to make a quick decision.
- From (2) we get ---------------------------------------------
(11 - n)! - (11 Choose n) = 330
where n is the number of Ascot party members selected.
Knowing GMAT/ETS, usually the results of (1) are aligned with the results of the (2). So in the above equation i started with n=4. And Voila LHS == RHS.
So n if 4. So we know that the number of Ascot Party members is not more than 4 .
Hence, I chose Answer D.
