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Problem of The Day (26.6.09)

with 6 comments


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

Written by Implex

June 26, 2009 at 4:25 am

6 Responses

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  1. sir
    do u provide solns for problem of the day??

    robbie

    June 27, 2009 at 3:05 am

  2. ans is b
    plz tell

    robbie

    June 27, 2009 at 3:39 am

  3. 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

  4. is it 216(B)??
    plz give da answer…

    sourav

    June 29, 2009 at 3:10 am

  5. Sorry, for being late in replies.

    The answer is option B) 216!

    Good luck !

    Rahul

    June 29, 2009 at 11:11 am

  6. can any1 explain how

    saurabh

    July 29, 2009 at 11:39 am


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