Number System
Aptitude

 Back to Questions
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

 Hide Ans

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
 ()

Smile you guys are awesome?

very very thank you.....