How many three digit numbers can be formed using the digits 1,2,3,4,5,6,7 and 8 without repeating the digits and such that the tens digit is greater than the hundreds digit and less than the units digit?
Answers
Here's the answer:
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Forming a 3 digit No. from A= {1,2,3,4,5,6,7,8}
Say, Hundreds digit = h
tens digit = t
and units digit = u
C be the condition that the
• The t>h & t<u
=> h < t < u
• The repetition of digits is not allowed.
{ °•° t>h and looking possible minimum value for t in A, we get t= 2 (min value).
when t= 1, h has no value satisfying Condition C from Set A.}
¶ Let see possible set of cases satisfying
{ h < t < u }
°{h}………………{t}……………{u}
• {1}………………{2}…{3,4,5,6,7,8}
• {1,2}……………{3}……{4,5,6,7,8}
• {1,2,3}…………{4}………{5,6,7,8}
• {1,2,3,4}………{5}…………{6,7,8}
• {1,2,3,4,5}……{6}……………{7,8}
• {1,2,3,4,5,6}…{7}………………{8}
Total No. of ways 'R' = n(h) + n(u) for every t={(2,3,4,5,6,7)}
R= (1×6) + (2×5) + (3×4) + (4×3) + (5×2) +(6×1)
R= 6 + 10 + 12 + 12 + 10 + 6 = 56 .
•°• The no. of 3 digit no.s that can be formed with C condition satisfied = 56
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:)
Hope it helps
Given : three digit numbers can be formed using the digits 1,2,3,4,5,6,7 and 8 without repeating the digit
Tens digit is greater than the hundreds digit and less than the units digit?
To Find : How many such numbers are possible
Solution:
Number is ABC
A < B < C
A = 1 B = 2 C can be in 6 ways ( 3 to 8)
A = 1 B = 3 C can be in 5 ways ( 4 to 8)
A = 1 B =7 C can be in 1 ways
Hence 6 + 5 + 4 + 3 + 2 + 1 = 21
if A = 2 Then 5 + 4 + 3 + 2 + 1 = 15
if A = 3 Then 4 + 3 + 2 + 1 = 10
if A = 4 Then 3 + 2 + 1 = 6 ways
if A = 5 then 2 + 1 = 3 ways
if A = 6 then 1 way
Hence total = 21 + 15 + 10 + 6 + 3 + 1 = 56
Hence 56 such numbers are possible
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