# Difficult Number System Solved QuestionAptitude Discussion

 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

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.

## (3) Comment(s)

Abhishek
()

The question says number has to be in acending order like 321,532 etc..

Now for any selection of three numbers from given 8 numbers there will be exactly one permutation that will follow that order... i.e suppose we select 3,7,9

Out of - 379,397,739,793,937,973 only 379 follows the order. Hence we will get exactly one number from every selection. i.e. 8C3 = 56 numbers

Mohit
()

ans don't include all the cases like 123, 267,

SUNIL
()

you guys are awesome?

very very thank you.....