What will be the sum of numbers?

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happy21

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I have a question:

A five digits number has to be formed by using the digits 1, 2, 3, 4 and 5 without repetition such that the even digits occupy odd places. Find the sum of all such possible numbers.

TIA.
 
happy21 said:
I have a question:

A five digits number has to be formed by using the digits 1, 2, 3, 4 and 5 without repetition such that the even digits occupy odd places. Find the sum of all such possible numbers.

TIA.
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
happy21 said:
A five digits number has to be formed by using the digits 1, 2, 3, 4 and 5 without repetition such that the even digits occupy odd places. Find the sum of all such possible numbers.
Click to expand...
Can you give one example of such a number? How do you define "odd places"?

It looks like you've given no thought at all to the question. That's part of the reason we ask to see your work.
 
happy21 said:
I have a question:

A five digits number has to be formed by using the digits 1, 2, 3, 4 and 5 without repetition such that the even digits occupy odd places. Find the sum of all such possible numbers.

TIA.
Click to expand...

Means even digits which are 2 and 4 (out of these five given numbers), and can come at position of unit's or hundreads' or ten-thousand's position (which are odd places when we consider the number from right to left...). Out of these three positions, at any two positions 2 and 4 will come in [MATH]\binom{3}{2}[/MATH] = 3 ways and 2 and 4 can be permuted in 2! ways. Remaining 3 digits (1, 3 & 5) will come in the remain positions in 3! ways. So in all there will be 3 x 2! x 3! = 36 numbers. And the sum of these 36 numbers is to be evaluated without simply adding all those numbers (which is not a method).....

Thx..
 
One of those 36 numbers will be 21435 (for example) because both 2 and 4 (which are even numbers) come at odd positions....Another example can be 53214...
 
happy21 said:
One of those 36 numbers will be 21435 (for example) because both 2 and 4 (which are even numbers) come at odd positions....Another example can be 53214...
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Okay, so the first thing we need to do is to take the wording literally, and not as "the even digits occupy all the odd places", which is a natural first impression. An example helps.

happy21 said:
Means even digits which are 2 and 4 (out of these five given numbers), and can come at position of unit's or hundreads' or ten-thousand's position (which are odd places when we consider the number from right to left...). Out of these three positions, at any two positions 2 and 4 will come in [MATH]\binom{3}{2}[/MATH] = 3 ways and 2 and 4 can be permuted in 2! ways. Remaining 3 digits (1, 3 & 5) will come in the remain positions in 3! ways. So in all there will be 3 x 2! x 3! = 36 numbers. And the sum of these 36 numbers is to be evaluated without simply adding all those numbers (which is not a method).....
Click to expand...
And the second thing is to know that you have indeed given it some thought, and that the hard part is to find the sum, not to count them, which you have done.

So now, as I understand it, you want to find the sum efficiently, without having to list them all. (You didn't make that a requirement in the OP -- though it definitely is a method, just not what you want.)

What we can do quickly is to just imagine we've made the list. What will that list look like?

I'm reminded of a similar problem: What is the sum of all permutations of the number 123? In the list of 6 numbers, each digit will appear in each place exactly twice, e.g. 123 and 132 both have 1 in the first place. So the sum will have 2(1+2+3) = 12 in each column, so the sum is 12 * 111 = 1332. I could do the same with more digits and save more work.

Do you see how to use that idea for this problem?
 
Dr.Peterson said:
Okay, so the first thing we need to do is to take the wording literally, and not as "the even digits occupy all the odd places", which is a natural first impression. An example helps.
Click to expand...

Even digits are only 2 but the odd places are 3, so all cannot be occupied (automatically it means this .....I believe)...

Dr.Peterson said:
So now, as I understand it, you want to find the sum efficiently, without having to list them all. (You didn't make that a requirement in the OP -- though it definitely is a method, just not what you want.)
Click to expand...

Yes....Actually I have not written all the numbers and not counted them all. Because if it were a problem with 9 digits, listing them all won't be an easy solution...so going with method will be logical....

Dr.Peterson said:
I'm reminded of a similar problem: What is the sum of all permutations of the number 123? In the list of 6 numbers, each digit will appear in each place exactly twice, e.g. 123 and 132 both have 1 in the first place. So the sum will have 2(1+2+3) = 12 in each column, so the sum is 12 * 111 = 1332. I could do the same with more digits and save more work.
Click to expand...

But here there is no additional condition imposed (e.g. even digits at odd places or like that...), so that's working easily....If I had to calculate the sum of all the 5 digit numbers formed with the digits 1, 2, 3, 4 & 5 (of course, without repetition of digits), without any other condition, I could have done that but only due to the additional constraint, I am trying to find the way....

Thx...
 
happy21 said:
But here there is no additional condition imposed (e.g. even digits at odd places or like that...), so that's working easily....If I had to calculate the sum of all the 5 digit numbers formed with the digits 1, 2, 3, 4 & 5 (of course, without repetition of digits), without any other condition, I could have done that but only due to the additional constraint, I am trying to find the way....
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Yes, the details are different, but I wanted to give a minimal hint suggesting a general idea, so you could enjoy discovering it. If I say too much, I'll give it away.

So, what will your list look like? Take your example 21435. Another number in the list will be 41235, with the two even digits swapped. Every number can be paired off with another that way. Then what can you do with the 135 part? Maybe something like what I did ...

happy21 said:
Even digits are only 2 but the odd places are 3, so all cannot be occupied (automatically it means this .....I believe)...
Click to expand...
My point was that the problem is easy to misread a problem like this, because something else seems more natural; my first thought was that either it was a typo (and the even digits should be in the even places), or it was a trick question (and the answer was zero, since there are no such numbers, if it meant "only"). In any case, it's important to think about what it means, to make sure you're thinking in the right direction (as it turns out you were).
 
What is the sum of all permutations of 21435 that leave 2 and 4 in the same pair of positions? This will use nothing more than I have said.

Then we can expand to the full problem.

There is also a different approach you could take that is faster; for that method you can look for a way to pair up numbers, in a way similar to the way we can add 1+2+3+...+100 as (1+100)+(2+99)+...+(50+51).

Please be aware that my goal is to let you do as much thinking as you can, in order to learn as much as possible, as well as enjoy the thrill of discovery. Don't give up.
 
Dr.Peterson said:
What is the sum of all permutations of 21435 that leave 2 and 4 in the same pair of positions?
Click to expand...
Its {[2 x (1 + 3 + 5) x (100 + 101 + 103)] + [6 x (2 x 104 + 4 x 102)]} = 140598.

Dr.Peterson said:
There is also a different approach you could take that is faster; for that method you can look for a way to pair up numbers, in a way similar to the way we can add 1+2+3+...+100 as (1+100)+(2+99)+...+(50+51).
Click to expand...

I am thinking this way...seems very interesting...I guess you mean like the position of 2 and 4 can be swaped (in our above example)....like that....

Dr.Peterson said:
Please be aware that my goal is to let you do as much thinking as you can, in order to learn as much as possible, as well as enjoy the thrill of discovery.
Click to expand...

Great...!
 
happy21 said:
Its {[2 x (1 + 3 + 5) x (100 + 101 + 103)] + [6 x (2 x 104 + 4 x 102)]} = 140598.
Click to expand...
Almost. I think you omitted the _4_2_ case in your second term.

happy21 said:
I am thinking this way...seems very interesting...I guess you mean like the position of 2 and 4 can be swaped (in our above example)....like that....
Click to expand...
There's more to it than that. Can you see a way to pair up every number in your list with another that will make a constant sum, as I did with pairs that make 101?
 
Dr.Peterson said:
What is the sum of 21435 and 45231?
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66666...Okay but how all such pairs can be easily detected....by different positions....may be...

Dr.Peterson said:
Now think. You need to be creative to solve problems like this; don't expect to be given a formula.
Click to expand...

I am not much experienced like you sir...Also I only have heard that knowledge increases when you distribute........At the same time you are true that I must be creative and try myself....

Thx.
 
happy21 said:
66666...Okay but how all such pairs can be easily detected....by different positions....may be...

I am not much experienced like you sir...Also I only have heard that knowledge increases when you distribute........At the same time you are true that I must be creative and try myself....
Click to expand...
As I said, THINK. Ask YOURSELF the question you asked here, before asking others.

How many numbers are there? Can each of them be paired up, or only some?

You can't learn how to think without doing it, any more than you can learn to throw or kick a ball without doing it for yourself. I am not just refusing to help; I am trying to help you learn. I also want to give you a chance to celebrate when you discover the answer!
 
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