q44

[Saumyojit]

How many 8-digit numbers are there the sum of whose digits is even?

Even All three diff & Even all three diff= 5c3 * 5c3 * 6!

Odd all three diff & Even all three alike : 5c3 * 5c1 * 6!/3!

Odd all three Alike & Even All three diff : 5c1 * 5c3 * 6!/3!


All 3 alike even and all three even alike , Two odd alike and one odd diff and two even alike one even diff , etc


.... If i solve this like this , then it would take a lot of time . Any other approach?

 

[pka]

How many 8-digit numbers are there the sum of whose digits is even?
Even All three diff & Even all three diff= 5c3 * 5c3 * 6!
Odd all three diff & Even all three alike : 5c3 * 5c1 * 6!/3!
Odd all three Alike & Even All three diff : 5c1 * 5c3 * 6!/3!
All 3 alike even and all three even alike , Two odd alike and one odd diff and two even alike one even diff , etc
.... If i solve this like this , then it would take a lot of time . Any other approach?

This seems to be yet another case-chaser. The lead digit cannot be zero. CORRECT?
In the above why do you consider three?
So the there are five odd digits and four even digits that can be the lead digit of an eight digit number.
For the sum to be even we need the number of odd digits to be even. WHY?
If all eight digits are even then there sum is even.
If all eight digits are odd then there sum is even. WHY?
That is the way that I would chase the cases.

 

[Saumyojit]

This seems to be yet another case-chaser. The lead digit cannot be zero. CORRECT?
In the above why do you consider three?
So the there are five odd digits and four even digits that can be the lead digit of an eight digit number.
For the sum to be even we need the number of odd digits to be even. WHY?
If all eight digits are even then there sum is even.
If all eight digits are odd then there sum is even. WHY?
That is the way that I would chase the cases.

q43

If the problem changes to: ten distinct transparent test tubes with diameter slightly larger than a ball, and 6 white and 6 black balls must go in them and at least 1 per tube, how many ways would the tubes appear? In this one your method would probably work. In our problem of urns, when you...

www.freemathhelp.com

Actually i could not understand where i was going wrong in this q43 post , so i was stressed out and made a mistake in writing the soln .

Yes it will be 8 digit ,
Odd + Odd = Even , Even + even= even

case 1: 8 even all alike = 4

2: 7 even all alike and one even diff: 4c1 * 4c1 * 7! / 6!

3:First spot any even except 0 and Seven places filled by only 0 = 4

4: Six even all alike & Two even another alike ; Six even all alike , Two even different (22222204) ; Five even different & Three even all same (20468000, 84602444) ...etc

If all eight digits are even then there sum is even.
If all eight digits are odd then there sum is even. WHY?

yes . But there will be so many diff cases .

This will take a lot of time like this

 

[pka]

yes . But there will be so many diff cases .

This will take a lot of time like this

As someone who has taught sections on counting many times, I am puzzles over all your similar questions on case chasseing.
If we begin the number with an even digit that is one of four. The other seven must contain an even number of odd digits,
I agree with you about the number of cases.

 

[Saumyojit]

As someone who has taught sections on counting many times, I am puzzles over all your similar questions on case chasseing.
If we begin the number with an even digit that is one of four. The other seven must contain an even number of odd digits,
I agree with you about the number of cases.

then how to do it

 

[pka]

then how to do it

Look here, I mean you no ill will, but I have already spent a great deal of time chasing down cases on your posts.
You should have some ideas on what needs doing. Remember that are multiple ways to do these.
If I were you, I would look at two overarching cases: eight digit numbers (i) lead digit odd & (ii) lead digit even.
For the sum of the digits to be even any odd digits must appear in pairs (tho not necessarily together).
Now you give it a go.

 

[sky2rain]

Basically the number of digits that is even doesn’t matter in this problem. The even sum solely depends on number of digits that is odd. If you have 3*5*7***, this pattern will always give you a odd sum if all 5 stars are even. You need a even number of odd digits to make the sum even.

I would imagine there are only 5 cases:

there’s 2 digit in all 8 are odd; 4 in all 8; 6 in all 8; 8 in all 8；0 in all 8.

once you calculate the total, exclude all 0 in front cases (starting with 8 0s would be easier).

I would also imagine that to get rid of the 0 in front is much more complicated than get the total of 8 digit permutations according to the problem requirement. Do you have a correct answer number on this problem yet?

 

[sky2rain]

Let me just do 1 case: 2 out of 8 are odd. It is basically 8 places to place the 2 odd numbers so 8c2. 5 choices for each odd number so 5^2. Then 5 choices for each of 6 even digit, that’s 5^6. So for this case you have 8c2*5^2*5^6=10937500. I don’t know why you would want to consider the repetition cases in this problem. The 5^2 or 5^6 or 5^whatever is a permutation with repetition for odd or even numbers from 0-9, so all the repetition cases are already included.

 

[sky2rain]

You can try to find all sum to be even for all 2 digit permutations and numbers and see if their ratio is 10:9. If they are it could be argued that the ratio for n-digit permutation with even sum to n-digit number with even sum is 10:9, and that saves you from excluding 0-in-front cases.

 

[pka]

If you had been in one of my classes I would tell you that I think that you lack the mathematical maturity to do these questions.
Now Prof Peterson may have a different take and bless him if so.
If an eight digit number has an even sum of its digits then it must contain [imath]0,~2,~4,~6.\text{ or }8[/imath] odd digits.
Any of these will work: [imath]83838383,~77777777,~42648093,\text{ or }22446688[/imath] and many others.

 

[Dr.Peterson]

How many 8-digit numbers are there the sum of whose digits is even?

I've been avoiding your posts lately because I dislike going back and forth with multiple people using different methods, and because they have required the "case chasing" pka commented on, which is simply not fun.

But please do as I've asked every other time, and show us the choices and the correct answer you were given. If nothing else, this helps us check if our answers make sense so we don't need to waste time trying to confirm.

This problem is actually very easy, and we've discussed an almost identical one before, namely here. (You never indicated whether you solved that one using my hint.) Your multiple cases are not needed this time, and are only distracting everyone. (I myself assumed it was needed until I decided to look only at the problem just now, and ignore everyone's work.)

 

[Saumyojit]

This problem is actually very easy, and we've discussed an almost identical one before, namely here.

45*10^6 is answer .

In the odd sum , it only comprised of combinations of odd + even

But here odd + odd and odd + even is there .

8 even all alike = 4

7 even all alike and one even diff: 4c1 * 4c1 * 7! / 6!

First spot any even except 0 and Seven places filled by only 0 = 4

Six even all alike & Two even another alike ; Six even all alike , Two even different (22222204) ; Five even different & Three even all same (20468000, 84602444), Four even alike & four even diff , four even alike and four 0's etc.....


There has to be a pattern that i am unable to find which will do this sum within small steps .

 

[Dr.Peterson]

I guess I have to tell you, since my hints don't get through.

How many 8-digit numbers are there the sum of whose digits is even?

If the sum of the first 7 digits is even, the last digit has to be even; there are 5 ways to do this.

If the sum of the first 7 digits is odd, the last digit has to be odd; there are 5 ways to do this.

So there are 9*10^6 ways to choose the first 7 digits, and 5 ways to choose the last, regardless of what the first 7 were; so the total is 9*10^6*5 = 45*10^6.

That's all it takes.

And most likely all the problems from this source can somehow be done without dozens of cases, if you think hard enough.

 

[Saumyojit]

9*10^6 ways to choose the first 7 digits, and 5 ways to choose the last, regardless of what the first 7 were; so the total is 9*10^6*5 = 45*10^6

yeah i see . Any even + even no of odds is mandatory to produce a even sum along with only 8 odds or no odds and only even digits .
I could have summed all the digits except zero for the no of ways to fill first spot which is 9 and then 10 * 10 ....
I see that last spot will always have 5 ways or choices to make both either one from the two sets according to the sum of first 7 digits .

If there is any odd no of odd digits upto 7th digit , then the last choice has to be from odd set .

Yes it covers all the cases .

 

[sky2rain]

0.9*10^8= 90*10^6, (90/2)*10^6=45*10^6, so for half of all 8-digit numbers, their sum is even.