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14 Jun 2019, 00:13
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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:
25%
(medium)
Question Stats:
79% (01:37) correct
21%
(01:45)
wrong
based on 178
sessions
History
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Time
Result
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How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)
A. 71
B. 200
C. 210
D. 840
E.1680
A. 71
B. 200
C. 210
D. 840
E.1680
14 Jun 2019, 00:19
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Bunuel
Let the odd digit be ABCD
Number of options for D = 4 (1,2,3 or 4)
A = 7, B = 6, C = 5
Total number of options = 7*6*5*4 = 840
IMO Option D
14 Jun 2019, 00:19
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Bunuel
Let us first take the units digit as there is a restriction there..
If the number is odd, the units digit can be any of 1, 3, 5 or 7 --- so 4 ways
The rest three digits can be in 7*6*5 ways..
Total ways --- ABCD = 7*6*5*4=840.... ABC in 7*6*5 ways and D in 4 ways
D
