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
sir
do u provide solns for problem of the day??
robbie
June 27, 2009 at 3:05 am
ans is b
plz tell
robbie
June 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
yodha
June 28, 2009 at 2:51 pm
is it 216(B)??
plz give da answer…
sourav
June 29, 2009 at 3:10 am
Sorry, for being late in replies.
The answer is option B) 216!
Good luck !
Rahul
June 29, 2009 at 11:11 am
can any1 explain how
saurabh
July 29, 2009 at 11:39 am