Permutations & Combinations - Find the sum

[zorro]

Homework Statement

Find the sum of all numbers greater than 10,000 formed by using the digits 0,2,4,6,8, no digit being repeated in any number.

The Attempt at a Solution

I can't think of any method of approach.
A hint will help.

 

[sjb-2812]

Homework Statement

Find the sum of all numbers greater than 10,000 formed by using the digits 0,2,4,6,8, no digit being repeated in any number.

The Attempt at a Solution

I can't think of any method of approach.
A hint will help.

How many numbers will have 2 as a first digit?

 

[zorro]

We have to find the sum of the numbers, not the number of numbers.

 

[sjb-2812]

Yes, I know that, but you wanted to know how we'd (I'd) go about solving it.

Don't forget if you have for instance 24 + 42 as a potential sum, you can swap the units column to give 22 + 44 which does not affect the sum.

 

[zorro]

I don't understand what you said.
How is that helpful here?

 

[sjb-2812]

If you know you have for instance 17 numbers that have 2 as a first digit, then you also know by symmetry that there are 17 with 4 as a first digit, with 6 as a first digit, with 8 as a first digit, and with 0 as a first digit. You can then shuffle the numbers around so you have 17 x 22222, 17 x 44444, 17 x 66666, 17 x 88888 (and 17 x 00000), which is a much easier sum to calculate.

 

[zorro]

Actually its solution is given in some other method. So I have a confusion between yours and this one-
Sum = (2+4+6+8)(4! x 10⁴) +(2+4+6+8)x 3 x 3! x (10³ + 10² + 10 +1)
Can you explain me this one?

 

[sjb-2812]

That's essentially the same thing as I have started explaining I think, though now I re-read the question I note my original answer also will count numbers like 02468 (which are not in your given range).

 

[zorro]

Ok then acc. to your method we have 24 nos. each starting with 2, 4, 6, 8
You should not take 0 as a first digit as any such number will be less than 10,000.
starting with 2, there are nos. like 24680, 24808 etc
how do you find their sum ? ( I really don't get why you took 17 x 22222, 17 x 44444 - all same digits as there can be different too)

 

[Tedjn]

Here is an explanation of the book solution. I want to calculate the sum

20468 + 20486 + ... + 24068 + ... + 42680 + ... + 86420.
​

This can be rewritten

(20000 + 400 + 60 + 8) + (20000 + 400 + 80 + 6) + ... + (20000 + 4000 + 60 + 8) + ... + (40000 + 2000 + 600 + 80) + ... + (80000 + 6000 + 400 + 20).
​

I can group all of the 10000 numbers together, all of the 1000 numbers together, all of the 100 numbers together, and so forth. The question then becomes:

How many 20000s do I have to add? Let's call this n. As argued above by symmetry, it's the same number as the number of 40000s and 60000s and 80000s I need to add. Therefore, the contribution to the sum by the 10000 numbers is (2+4+6+8)*10000*n. Repeat for the 1000 numbers, the 100 numbers, and so forth.

Do you see how this matches the book's solution method?

 

[zorro]

Thanks for your explanation.
I understood how this came (2+4+6+8)(4! x 10⁴)
Now the next terms should be (2+4+6+8)(3! x 10³) + (2+4+6+8)(2! x 10²) + (2+4+6+8)(1! x 10) + (2+4+6+8)

This is my explanation - the terms with 10³ can be arranged in 3! ways as no repetition is allowed.
The same explanation with 2! 1! follows.
Why did he take 3! x 3 in common for next terms?

 

[sjb-2812]

Thanks Tedjn, I have been trying to write out my thoughts clearly, but you seem to have done a good job of working out what I meant.

I think the 3! x 3 (and similar) terms take into account that the thousands, hundreds, tens and units digits can be zero; as well as 2, 4, 6 and 8.

Note that n! = n * (n-1)! = (n-1+1) * (n-1)! = (n-1)[(n-1)!] + (n-1)! (though not sure if this really helps)