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 = \(5^7-5^6\)

There 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 = \(C^7_3*5^3*5^4-C^6_2*5^2*5^4\)

We 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 = \(C^7_1*5^1*5^6-5^6\)

We 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=

\(5^7-5^6\)+\(C^7_5*5^7-C^6_4*5^6\)+ \(C^7_3*5^7-C^6_2*5^6\)+\(C^7_1*5^7-5^6\)

=\(4*5^6+90*5^6+160*5^6+34*5^6\)

=\(288*5^6\)

Zareentaj - What's the OA ?

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