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29 May 2018, 01:54
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
86% (01:45) correct
14%
(01:39)
wrong
based on 297
sessions
History
Date
Time
Result
Not Attempted Yet
A 3-character alpha-numeric code does have the following properties – the first character can be any number except 0 and 9, the second character can be any small letter between a to z, excluded, and the third character can have any of those characters possible for the first two places. How many such codes can be formed?
A. 26
B. 64
C. 520
D. 6144
E. 9360
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Article-1: Learn when to “Add” and “Multiply” in Permutation & Combination questions
Article-2: Fool-proof method to Differentiate between Permutation & Combination Questions
Article-3: 3 deadly mistakes you must avoid in Permutation & Combination
All Articles: Must Read Articles and Practice Questions to score Q51 !!!!
A. 26
B. 64
C. 520
D. 6144
E. 9360
Related Articles:
Article-1: Learn when to “Add” and “Multiply” in Permutation & Combination questions
Article-2: Fool-proof method to Differentiate between Permutation & Combination Questions
Article-3: 3 deadly mistakes you must avoid in Permutation & Combination
All Articles: Must Read Articles and Practice Questions to score Q51 !!!!
29 May 2018, 03:04
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EgmatQuantExpert
The 3 character alpha-numeric code has the following characters
First character - Any number except 0 and 9 - Total of 10 - 2 = 8 possibilities
Second character - Any small letter except a and z - Total of 26 - 2 = 24 possibilities
Third character - Any number except 0 and 9 or Any small letter except a and z - Total of 8 + 24 = 32 possibilities
In this case, to find the total number of possibilities, we can use the unit's digit to find the
answer as number ending 8 * number ending 4 * number ending 2 is a number ending 4.
Also, the answer must be close to 10*20*30 is a little more than 6000.
P.S when the answer options are far apart, we can use approximations to arrive at the answer
Therefore, Option D(6144) is the the total number of 3-character alpha-numberic codes.
29 May 2018, 03:32
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pushpitkc
I think your calculations and answer is correct but by mistake you have mentioned option C in your answer (in brackets though the actual answer is right). I hope this helps.
