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Saumyojit

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find the number of seven-digit numbers which can be formed with the sum of the digits being odd.​

my approach is breaking down all the cases : ​

Case 1 : first digit is odd --> 5waysEg: 1000000 etc ​

Case 2: Odd + even : 5c1 * 4c1 *2 = 40 way Eg :1200000 , 1400000

Case 3 : One odd + Even repeated twice : 5c1 * 4c1 *3 = 60 Eg:1220000 , 221000 eg ..

Case 4 : One odd + Even repeated thrice : 5c1 * 4c1 *4= 80 Eg:1222000 , 221200 eg ..
But this way will take time . Please give shortcut
 
Here's a hint:

Suppose I pick any values I want for the first 6 digits; for example, maybe I pick 380493_.

How many choices do I have for the seventh digit?
 
Dr.Peterson said:
Suppose I pick any values I want for the first 6 digits; for example, maybe I pick 380493_.

How many choices do I have for the seventh digit?
Click to expand...
That will depend what is the sum of first six digits if it is odd then zero or even digit and if it is even then it has to be odd .
So , what will be the approach ?
 
Saumyojit said:
That will depend what is the sum of first six digits if it is odd then zero or even digit and if it is even then it has to be odd .
So , what will be the approach ?
Click to expand...
Okay, then, consider each of these two cases. If the sum of the first six digits is odd, how many choices are there for the seventh? Then repeat if it's even.

A lesson I have repeatedly tried to teach is: TRY something; don't just look at what you have and give up. Sometimes you will discover that it is easier than you think! This is especially true if someone with experience has TOLD you to try it.

By the way, "zero or even digit" is not necessary; zero is even!
 
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Dr.Peterson said:
Okay, then, consider each of these two cases. If the sum of the first six digits is odd, how many choices are there for the seventh? Then repeat if it's even.

A lesson I have repeatedly tried to teach is: TRY something; don't just look at what you have and give up. Sometimes you will discover that it is easier than you think! This is especially true if someone with experience has TOLD you to try it.

By the way, "zero or even digit" is not necessary; zero is even!
Click to expand...
1odd + 6even , 3odd + 4even , 5odd + 2ven , all 7 odd
 
You're ignoring my hint! You only need two cases: Sum of first six digits odd/last even; sum of first six digits even/last odd. And these two cases reduce to only one case, with the result that you don't need to count either case individually.

Do it the hard way if you want, but I thought you were here to learn new things. I won't try to help you with hard ways.
 
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