Accenture
Company
Numerical Ability
Permutation and Combination
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
A. 5 B. 10
C. 15 D. 20
Read Solution (Total 4)
-
- last position is fixed for 5.
so for 1st position = 5 choices
2nd position= 4 choices.
hence, 5*4*1=20 - 12 years agoHelpfull: Yes(20) No(3)
- there will be 3 position .the last position will be fixd because 5 is only divided by 0 and 5.so the last position will be fixed for the 5.now for 1st 2 position it will be 5p2=20.
ans=20 - 10 years agoHelpfull: Yes(7) No(3)
- its 20
as the no. has to be divisible by 5, the unit digit is 5, now we have 5 no.s which have to be at tensd and hundreds place and 5 numbers can be chosen for 2 places in 5*4 i.e, 20 ways. - 9 years agoHelpfull: Yes(0) No(1)
- answer is 20
- 7 years agoHelpfull: Yes(0) No(0)
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