# Thread: A five-digit number is formed using digits 1, 3, 5, 7 and 9 .......

1. ## A five-digit number is formed using digits 1, 3, 5, 7 and 9 .......

A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
I know that someone has posted the solution to this but I cant understand it. Can anyone please post the answer by explaining all the details. I have recently started learning permutation and combinations. Please simplify the answer as much as you can. Thank you.  Reply With Quote

2. Hi senoritace,

I'll try to be as detailed as possible, but if I'm not clear enough about something, please let me know and I'll try to do a better job. (In trying to be detailed I also might end up explaining some things that you already know).

Ok, so we have 5 different values we're working with (1, 3, 5, 7 and 9). The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits (i.e. the number of 'permutations').

Here's one way to calculate this number:

For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 5*4*3*2*1 (commonly written 5!) = 120 different 5-digit numbers.

Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.

120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...

This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.

We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 5*10,000, the 3 is 3*1000, etc...

So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:

Sum of ten thousands digits: 24*10,000*(1 + 3 + 5 + 7 + 9)
+
Sum of thousands digits: 24*1,000*(1 + 3 + 5 + 7 + 9)
+
Sum of hundreds digits: 24*100*(1 + 3 + 5 + 7 + 9)
+
Sum of tens digits: 24*10*(1 + 3 + 5 + 7 + 9)
+
Sum of units digits: 24*1*(1 + 3 + 5 + 7 + 9)

= 6,666,600

I hope that helps! Feel free to ask for clarification if I've been unclear about anything, these can definitely be confusing problems!  Reply With Quote

3. very helpful suggestion for this kind of problem....  Reply With Quote

4. ## thank u very much. i got it. thanks for the simple effective explanation. Originally Posted by thepillow Hi senoritace,

I'll try to be as detailed as possible, but if I'm not clear enough about something, please let me know and I'll try to do a better job. (In trying to be detailed I also might end up explaining some things that you already know).

Ok, so we have 5 different values we're working with (1, 3, 5, 7 and 9). The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits (i.e. the number of 'permutations').

Here's one way to calculate this number:

For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 5*4*3*2*1 (commonly written 5!) = 120 different 5-digit numbers.

Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.

120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...

This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.

We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 5*10,000, the 3 is 3*1000, etc...

So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:

Sum of ten thousands digits: 24*10,000*(1 + 3 + 5 + 7 + 9)
+
Sum of thousands digits: 24*1,000*(1 + 3 + 5 + 7 + 9)
+
Sum of hundreds digits: 24*100*(1 + 3 + 5 + 7 + 9)
+
Sum of tens digits: 24*10*(1 + 3 + 5 + 7 + 9)
+
Sum of units digits: 24*1*(1 + 3 + 5 + 7 + 9)

= 6,666,600

I hope that helps! Feel free to ask for clarification if I've been unclear about anything, these can definitely be confusing problems!

thanks. I got the solution now. you have explained it nicely. thanks.  Reply With Quote

5. Originally Posted by thepillow Hi senoritace,

I'll try to be as detailed as possible, but if I'm not clear enough about something, please let me know and I'll try to do a better job. (In trying to be detailed I also might end up explaining some things that you already know).

Ok, so we have 5 different values we're working with (1, 3, 5, 7 and 9). The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits (i.e. the number of 'permutations').

Here's one way to calculate this number:

For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 5*4*3*2*1 (commonly written 5!) = 120 different 5-digit numbers.

Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.

120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...

This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.

We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 5*10,000, the 3 is 3*1000, etc...

So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:

Sum of ten thousands digits: 24*10,000*(1 + 3 + 5 + 7 + 9)
+
Sum of thousands digits: 24*1,000*(1 + 3 + 5 + 7 + 9)
+
Sum of hundreds digits: 24*100*(1 + 3 + 5 + 7 + 9)
+
Sum of tens digits: 24*10*(1 + 3 + 5 + 7 + 9)
+
Sum of units digits: 24*1*(1 + 3 + 5 + 7 + 9)

= 6,666,600

I hope that helps! Feel free to ask for clarification if I've been unclear about anything, these can definitely be confusing problems!
You have multiplies every set by 24... Well we have 24 possibilities when we have a digit locked at the ten thousandth spot. For example
If we have 1_ _ _ _ so now we have 4 sponts left and a total of 4*3*2*1 possiblities. Also we have considered/added all 10 thousands possible.
then when we move on to the thousandth spot and say we plug 1 as--- _ 1 _ _ _, so wouldn't we only consider the hundred, thens and units as 3*2*1. I hope my point is clear.

My solution
10,000*24*(1+3+5+7+9) = 6,000,000
1,000*6*(1+3+5+7+9)=150,000
100*2*(1+3+5+7+9)=5,000
10*1*(1+3+5+7+9)=250
1*1*(1+3+5+7+9)=25
Total Sum = 6,155,275  Reply With Quote

6. Originally Posted by Fahad Ali You have multiplies every set by 24... Well we have 24 possibilities when we have a digit locked at the ten thousandth spot. For example
If we have 1_ _ _ _ so now we have 4 sponts left and a total of 4*3*2*1 possiblities. Also we have considered/added all 10 thousands possible.
then when we move on to the thousandth spot and say we plug 1 as--- _ 1 _ _ _, so wouldn't we only consider the hundred, thens and units as 3*2*1. I hope my point is clear.

My solution
10,000*24*(1+3+5+7+9) = 6,000,000
1,000*6*(1+3+5+7+9)=150,000
100*2*(1+3+5+7+9)=5,000
10*1*(1+3+5+7+9)=250
1*1*(1+3+5+7+9)=25
Total Sum = 6,155,275

Hey Fahad, the reason I multiply by 24 is because this is the number of times that each of our 5 numbers will appear in the units place, the tens place, hundreds place, etc.

Maybe this will make it a bit clearer:

Here's an abridged list of the numbers we get when rearranging our digits:

13579
19537
17395
...
31597
39157
...
59317
57931
...
79315
73159
...
93517
97531

These are just a few of the numbers, but if you listed all of them out you'd see that you would have 120 numbers. 24 of them would start with 1, 24 would start with 3, 24 would start with 5, etc. 24 would also have 1 as the second digit, 24 would have 3 as the second digit, etc. This is true for all the number and all the digits. This is because if we take any of our 5 numbers and pick which digit it will be, then there are 4! = 24 ways to rearrange the other numbers around it (regardless of which position it's in).

Does that make sense?  Reply With Quote

7.  Reply With Quote