zareentaj wrote:

How many different 7 digit members are their sum of whose digits is even ?

Here is how i approached it.

Sum of the 7 digits is even if the number has 1 or 3 or 5 or 7 even digits

There are 5 even digits available 0,2,4,6,8 and 5 odd digits available 1,3,5,7,9. However the first digit in the 7 digit number should not be 0. Otherwise it will become a 6 digit number.

7E 0O =

5^7-5^6There are 5 even numbers available. So each of 7 the digits can be filled with an even number in

5^7 ways, but this also contains the numbers having 0 in the first digit location.

If first digit has a 0 , rest 6 digits can be filled with an even number in

5^6 ways

5E 2O =

C^7_5*5^5*5^2-C^6_4*5^4*5^2 We can first pick the 5 spots having even numbers in

C^7_5 ways ,fill them each with even number in

5^5 ways and fill the remaining 2 digits with odd numbers in

5^2 ways. Again we need to subtract the numbers having 0 in the first digit location

If first digit has a 0 , we can pick the 4 spots out of the remaining 6 having even numbers in

C^6_4 ways ,fill them each with even number in

5^4 ways and fill the remaining 2 digits with odd numbers in

5^2 ways.

3E 4O =

C^7_3*5^3*5^4-C^6_2*5^2*5^4We can first pick the 3 spots having even numbers in

C^7_3 ways ,fill them each with even number in

5^3 ways and fill the remaining 4 digits with odd numbers in

5^4 ways. Again we need to subtract the numbers having 0 in the first digit location

If first digit has a 0 , we can pick the 2 spots out of the remaining 6 having even numbers in

C^6_2 ways ,fill them each with even number in 5^2 ways and fill the remaining 4 digits with odd numbers in

5^4 ways.

1E 6O =

C^7_1*5^1*5^6-5^6We can first pick the 1 spot having an even number in

C^7_1 ways ,fill it with an even number in

5^1 ways and fill the remaining 6 digits with odd numbers in

5^6 ways. Again we need to subtract the numbers having 0 in the first digit location

If first digit has a 0 , we can fill the remaining 6 digits with odd numbers in

5^6 ways.

Total no of 7 digit numbers whose sum of digits is even=

5^7-5^6+

C^7_5*5^7-C^6_4*5^6+

C^7_3*5^7-C^6_2*5^6+

C^7_1*5^7-5^6=

4*5^6+90*5^6+160*5^6+34*5^6=

288*5^6Zareentaj - What's the OA ?

Kudos for the patience displayed. Although if taken this approach in GMAT I wonder if it can be done in 2 mins? Assume Bunuel or Ian's approach is what they would be looking for...