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Difficulty:
65%
(hard)
Question Stats:
67% (02:39) correct
33%
(02:09)
wrong
based on 132
sessions
History
Date
Time
Result
Not Attempted Yet
Number of four digit numbers with all digits different and containing the digit 7 is
(A) 2016
(B) 1884
(C) 1848
(D) 1828
(E) 1824
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(A) 2016
(B) 1884
(C) 1848
(D) 1828
(E) 1824
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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Bunuel
There are few different ways to answer this question.
One option is to consider for cases:
- 4-digit numbers in the form 7XXX
- 4-digit numbers in the form X7XX
- 4-digit numbers in the form XX7X
- 4-digit numbers in the form XXX7
The number of 4-digit numbers in the form 7XXX = (1)(9)(8)(7) = 504
The number of 4-digit numbers in the form X7XX = (8)(1)(8)(7) = 448 [ the first digit can be any digit other than 0 or 7]
The number of 4-digit numbers in the form XX7X = (8)(8)(1)(7) = 448 [ the first digit can be any digit other than 0 or 7]
The number of 4-digit numbers in the form XXX7 = (8)(8)(7)(1) = 448 [ the first digit can be any digit other than 0 or 7]
504 + 448 + 448 + 448 = 1848
Answer: C
General Discussion
Bunuel
Total ways to choose four digits, in which one is 7 => 1C1*9C3=84
Each way can be arranged in 4! ways => 84*4! = 2016
But these contain numbers where 0 is first digit.
If 0 and 7 are already selected, ways to select remaining 2 is 8C2 or 28 ways.
Arrangement where 0 is already placed in first position is 3! ways => 28*3! Or 168
Ways = 2016-168 = 1848
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