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10 May 2020, 23:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (01:49) correct
26%
(02:25)
wrong
based on 80
sessions
History
Date
Time
Result
Not Attempted Yet
How many numbers not exceeding 10000 can be made using the digits 2, 4, 5, 6, 8 if repetition of digits is allowed?
A. 740
B. 940
C. 780
D. 820
E. 660
A. 740
B. 940
C. 780
D. 820
E. 660
11 May 2020, 00:07
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1-digit numbers = 5
2-digit numbers = 5*5=25
3-digit numbers = 5*5*5=125
4-digit numbers = 5*5*5*5=625
Total possible numbers = 5+25+125+625 = 780
2-digit numbers = 5*5=25
3-digit numbers = 5*5*5=125
4-digit numbers = 5*5*5*5=625
Total possible numbers = 5+25+125+625 = 780
preetamsaha
11 May 2020, 00:31
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preetamsaha
Another way to realize the way combination can be played around with. The correct and easy way would be as shown above.
The words can be, 1-, 2-. 3- or 4-digit....
ABCD
If I take A's possible value as 0 too, I will get both 3- and 4-digit number...
So A can be any of 6 ( 0, 2, 4, 5, 6, 8), and remaining 3 places can be filled by any 5 (2, 4, 5, 6, 8) => 6*5*5*5=750
Now we are left with 1- and 2- digit numbers ..XY
If I take X's possible value as 0 too, I will get both 1- and 2-digit number...
So X can be any of 6 and B any of 5 => 6*5=30
Total 750+30=780
