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Difficulty:
75%
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Question Stats:
58% (03:07) correct
42%
(02:47)
wrong
based on 129
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How many three-digit positive integers, with digits a, b and c exist such that a < b and c < b? Assume that a is in hundred's place, b is in ten's place, c is in unit's place and a is a non-zero digit.
A. 120
B. 240
C. 364
D. 450
E. 648
A. 120
B. 240
C. 364
D. 450
E. 648
04 Jan 2020, 00:23
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b=2; a=1; c= 0 or 1; Total possible numbers= 1*2
b=2; a=1 or 2; c= 0, 1 or 2; Total possible numbers= 2*3
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b=8; a=1,2....., 7; c= 0, 1........, 7; Total possible numbers= 7*8
b=9; a=1,2....., 8; c= 0, 1........, 8; Total possible numbers= 8*9
Total possible numbers= (1*2)+(2*3)+......+(7*8)+(8*9)
\(Σn*(n+1)= Σ(n^2+n)\)= \(\frac{n*(n+1)*(2n+1)}{6}+\frac{n*(n+1)}{2}\)
Put n=8, we get (1*2)+(2*3)+......+(7*8)+(8*9)= 240
b=2; a=1 or 2; c= 0, 1 or 2; Total possible numbers= 2*3
.
.
.
b=8; a=1,2....., 7; c= 0, 1........, 7; Total possible numbers= 7*8
b=9; a=1,2....., 8; c= 0, 1........, 8; Total possible numbers= 8*9
Total possible numbers= (1*2)+(2*3)+......+(7*8)+(8*9)
\(Σn*(n+1)= Σ(n^2+n)\)= \(\frac{n*(n+1)*(2n+1)}{6}+\frac{n*(n+1)}{2}\)
Put n=8, we get (1*2)+(2*3)+......+(7*8)+(8*9)= 240
DisciplinedPrep
