## Problem of The Day (26.6.09)

**Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5,
using each digit exactly once such that exactly two odd positions are occupied
by odd digits. What is the sum of the digits in the rightmost position
of the numbers in S?
a. 228 b. 216 c. 294 d. 192 e. None of these
**

Advertisements

sir

do u provide solns for problem of the day??

robbieJune 27, 2009 at 3:05 am

ans is b

plz tell

robbieJune 27, 2009 at 3:39 am

i m also getting (e).

_ _ _ _ _

1 2 3 4 5

odd even

if we fix one odd and one even no. at 2nd and 4th position,

2odds and 1 even are left, which have to be filled up at the odd places.

i.e. 3C1 * 2C1 * 2! * 3! = 72.

i.e. 12 * (1+2+3+4+5) = 180

yodhaJune 28, 2009 at 2:51 pm

is it 216(B)??

plz give da answer…

souravJune 29, 2009 at 3:10 am

Sorry, for being late in replies.

The answer is option B) 216!

Good luck !

RahulJune 29, 2009 at 11:11 am

can any1 explain how

saurabhJuly 29, 2009 at 11:39 am