Number System
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Q.

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?

 A.

48

 B.

56

 C.

64

 D.

72

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Solution:
Option(B) is correct

Ten's digit = 7 ⇒ units digit =8 ⇒ Hundred's digit $=1,2,3,4,5,6.$
⇒ Number of ways $=1×6$

Ten's digit = 6 ⇒ units digit $=7, 8$ ⇒ Hundred's digit $=1,2,3,4,5.$
⇒ Number of ways $=2×5$

Ten's digit = 5 ⇒ units digit $=6,7,8$ ⇒ Hundred's digit $=1,2,3,4$
⇒ Number of ways $=3×4$

Ten's digit = 4 ⇒ units digit $=5, 6, 7, 8$ ⇒ Hundred's digit $=1,2,3$.
⇒ Number of ways $=4×3$

Ten's digit = 3 ⇒ units digit $=4,5,6,78$ ⇒ Hundred's digit $=1,2$
⇒ Number of ways $=5×2$

Ten's digit = 2 ⇒ units digit $=3,4,5,6,7,8$ ⇒ Hundred's digit $=1$
⇒ Number of ways $=6×1$

Total number of ways $= 6+10+12+12+10+6 =$ 56.


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