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26 Aug 2022, 23:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:13) correct
43%
(02:38)
wrong
based on 108
sessions
History
Date
Time
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Not Attempted Yet
A firm has 20 divisions, and each division has been assigned a code using letters (alphabets). If a code is formed by two or three distinct letters and if one division is represented by only a single code – either two or three letters, then what is the minimum number of letters needed to assign codes to all these divisions? (Assume that the order of the letters in the code does not matter – so AB and BA mean the same code)
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
27 Aug 2022, 03:08
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Snehaaaaa
Say there are minimum of \(n\) letters needed, then;
The number of pair of distinct letter codes possible would be \(C^2_n\);
The number of three distinct letter codes possible would be \(C^3_n\);
We want \(C^2_n+C^3_n\geq{20}\):
- \(\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{3!}\geq{20}\);
\(3n(n-1)+n(n-1)(n-2)\geq{60}\);
\(n^3 - n\geq{60}\);
\(n\geq{5}\);
\(n_{min}=5\).
Answer: C.
Or else one could just test numbers.
- Four letters are enough for \(C^2_4+C^3_4=6+4=10\) codes. Not enough.
Five letters are enough for \(C^2_5+C^3_5=10+10=20\) codes. Good!
Hope it's clear.
P.S. Here is an official questions this one is based on: https://gmatclub.com/forum/a-researcher ... 34584.html
12 Oct 2023, 01:38
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