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sritamasia
27 Nov 2021, 14:46
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
36% (01:53) correct
64%
(02:12)
wrong
based on 131
sessions
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Not Attempted Yet
The number of ways of inviting at least one executive out of \(N\) executives to a conference is \(X\).
What is \(N\)?
(1) \(3<N<7\)
(2) \(15<X<63\)
What is \(N\)?
(1) \(3<N<7\)
(2) \(15<X<63\)
TryingToAceVerbal
Joined: 18 Nov 2021
Last visit: 07 Aug 2022
Posts: 13
Given Kudos: 285
Location: Germany
Concentration: Finance, Strategy
Schools: Harvard '24 Wharton '24 Stanford '24
GMAT 1: 620 Q50 V25
GMAT 2: 700 Q50 V36
01 Dec 2021, 07:50
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sritamasia
My approach:
\(X_{y} =\) number of ways of inviting Y people out of N executives
If all N guests are invited:
\(X_{N} = N * (N-1) * (N-2) * ... * 1 / N! = N! / (0! * N!) = 1\)
If N-1 guests are invited:
\(X_{N-1} = N * (N-1) * (N-2) * ... * 2 / (N-1)! = N! / (1! * (N-1)!) = N! / (N-1)!\)
If N-2 guests are invited:
\(X_{N-2} = N * (N-1) * (N-2) * ... * 3 / (N-2)! = N! / (2! * (N-2)!) = N! / (2 * (N-2)!)\)
...
If 1 guest is invited:
\(X_{1} = N\)
\(=> X = X_{N} + X_{N-1} + ... + X_{1}\)
(1) N could be 4, 5, or 6. No further restriction.
=> not sufficient
(2) \(15<X<63\)
Let's try some numbers:
case 1: \(N = 4\)
\(X = 4! / 4! + 4! / 3! + 4*3 / 2! + 4 / 1! = 1 + 4 + 6 + 4 = 15\)
But X is required to be bigger than 15
case 2: \(N = 5\)
\(X = 5! / 5! + 5! / 4! + 5*4*3 / 3! + 5*4 / 2! + 5 / 1! = 1 + 5 + 10 + 10 + 5 = 31\)
X could be 31, because \(15<31<63\)
case 3: \(N = 6\)
\(X = 6! / 6! + 6! / 5! + 6*5*4*3 / 4! + 6*5*4 / 3! + 6*5 / 2! + 6 / 1! = 1 + 6 + 15 + 20 + 15 + 6 = 63\)
But X is required to be smaller than 63
=> \(X = 31\) and therefore \(N = 5\)
=> B
02 Dec 2021, 02:53
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Number of ways are 2^N. Please correct me if I am wrong.
