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Differential Calculus

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28 Dec 2011 22:50:39 IST

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permutations

Mathematics

the number of all 5 digited numbers that can be made the sum of whose digits is even is ________________

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rohit nimmagadda

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29 Dec 2011 22:40:25 IST

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@NAGEEN: actually i didnt solve the problem completely but i can tell u a method to do the problem..... the possile 5 digited no.'s will be in this way case-1 : talking all the digits as even case-2 : taking 3 even and 2 odd no.'s case-3 taking 1 even and 4 odd the nol. of permutations for the first case is 4*5*5*5*5 bcoz we have 5 even no.'s in 0 to 9 so we can use any of them for each odd so the no. of choices for the 2nd,3rd,4th,5th digits are 5 but in the first digit 0 should not be taken so the choices are 4 so the total permutations are 4*5*5*5*5 for case-2 : arrangemens are posssible b/w even's and odd's so we first take those arrangements and repeat the same procedure as done in case1 ..... can u tell me the ans. so that i can tell u if get the ans. by tomorrow

nageen

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30 Dec 2011 21:30:01 IST

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i got its solution rohit...the first four digits of the no. can b taken in 9*10*10*10 ways.now sum of the first four digits is either even or odd.if the sum is even the fifth digit is either 0,2,4,6 or 8 or if the sum is odd then the last digit can b either1,3,5,7 or 9 i.e the last digit can b taken in 5 ways the ans is 9*10*10*10*5

rohit rajput

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5 Jan 2012 22:15:16 IST

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If repition is not allowed...then

(9*9*8*7*6)/2 because even terms will be half the total terms

=13608.

if repition allowed then

(9*10*10*10*10)/2.......for same reasons.

=45000.

Arjun Virmani

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Joined: 1 Jun 2009

Posts: 1294

7 Jan 2012 05:26:51 IST

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The last digit trick works best!!

ans 9*10*10*10*5 !!!

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