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04 Feb 2021, 21:40
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B
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Difficulty:
75%
(hard)
Question Stats:
59% (02:58) correct
41%
(02:27)
wrong
based on 188
sessions
History
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Not Attempted Yet
How many 3-digit positive integers, with digits a, b, and c exist such that a < b and c < b? Assume that a is in the hundred's place, b is in ten's place, c is in the unit's place and a is a non-zero digit.
A. 100
B. 240
C. 364
D. 450
E. 648
A. 100
B. 240
C. 364
D. 450
E. 648
04 Feb 2021, 22:58
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DisciplinedPrep
The number is abc, and the other two variables, a and c, are dependent on the value of b.
Minimum value of a = 1, so b has to be greater than 1.
1) b=2, then a can be only 1, and c can be 0 or 1. Total numbers = 1*1*2=2
2) b=3, then a can be 1 or 2, and c can be 0, 1, or 2. Total = 2*1*3=6
So, value that a takes is one less than the numeric value of b, and value that b takes is equal to the numeric value of b.
Thus when b=9, a can take 8 values and c can take 9 values = 8*1*9=72
Total = \(1*1*2+2*1*3+3*1*4+4*1*5+5*1*6+6*1*7+7*1*8+8*1*9=2+6+12+20+30+42+56+72=240\)
B
General Discussion
05 Feb 2021, 02:49
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DisciplinedPrep
There are 3 ways of getting the number:
1. a, b and c could be all distinct with no 0. e.g. 132, 231
Pick 3 distinct digits from 1 - 9 in 9C3 ways and arrange them in 2 ways. The greatest one in the middle and other two around it.
2. a, b and c all distinct with c = 0. e.g. 230, 590 etc
Pick 0 and pick 2 digits from 1 - 9 in 9C2 ways. There is only 1 way of arranging the digits.
3. a and c are same and b greater. e.g. 131
Pick 2 digits in 9C2 ways and there is one 1 way of arranging.
Total = 9C3*2 + 9C2 + 9C2 = 240
Answer (B)
