How many five digit integers can we make chosen our digits from 5,8,8, 2, 6, 6, 0?

Question

as my last question here this is the same problem but with 2 repeated digits,

here is my approach:

To solve this problem we need to devide it into 5 cases:

1 The integer contains one 8 and one 6 : 4 x 4 x 3 x 2 x 1 = 96

2- The integer contains one 8 and two 6s or one 6 and two 8s and no 0: (5C2) x 3 x 2 x 1 = 60

3- As same as 2 but containing 0: (4C1) x (4C2) x (3C2) x 2 = 144

4- The integer contains two 8’s and two 6’s with no 0: (5C2) x (3C2) x (2C1) x 2 = 120

5- Same as 4 with 0: (4C1) x (4C2) x (2C2) = 24

The final result would be : 96 + 2(60) + 2(144) + 120 + 24 = 648

but when I used an online tool I got 588 as result.

(ii)Find the number of integers in (i) which satisfies “If there is 8, then it is followed by 6”

This is a FAQ and something you should first examine from literature on your own. There's no much sense in explaining something you should better conclude by yourself (for your own good). Furthermore, you can compare the older posts with your approach. I believe there's more benefit from that. — PinkyWay, Jun 03 '20 at 22:45
Also, 3 months ago, other users told you to learn how to format the questions. Your post is unreadable. — PinkyWay, Jun 03 '20 at 22:48
See: How many distinct four-digit integers can one make from the digits $1,3,3,7,7,8$?. This is obviously the same question. — PinkyWay, Jun 03 '20 at 22:50
Another one: Given an alphabet with 6 non-distinct integers, how many distinct 4-digit integers are there?. — PinkyWay, Jun 03 '20 at 22:52
One more: permutations with repeating symbols and binomial coefficient. — PinkyWay, Jun 03 '20 at 22:54
Does this answer your question? How many distinct four-digit integers can one make from the digits $1,3,3,7,7,8$? — PinkyWay, Jun 03 '20 at 23:00
@Cheesecake thanks for your notes I will consider it while posting again, I am so sorry for bad format in math writing it's new for me, I did read these posts and my question is why the computer program outputs 588 which is different from my result I understood the methods in these questions and applied it and I didn't got the 588 which made me ask about why is it 588 thetool , sorry again. — D0mBas3, Jun 04 '20 at 14:02

score 1 · Answer 1 · answered Jun 03 '20 at 23:11

There are $7$ numbers to choose from, $\{ 5,8,8,2,6,6,0 \}$. I assume that an integer should not start with $0$

There are $7!$ possible permutations of the numbers, but only $6/7$'s of those begin with a non-zero number, so $6(6!)$ permutations. We have overcounted these permutations by a factor of $2$ since the $8$'s are interchangeable. We have also overcounted by a factor of $2$ for the interchangeable $6$'s. So far we are at $$ \frac{6(6!)}{(2!)(2!)} $$ permutations. For each permutation, the first five digits form an integer, but it does not matter what order the last two digits are in. So we have overcounted by a factor of $2$. We arrive at the final answer of $$ \frac{6(6!)}{(2!)(2!)(2!)} $$ possible five-digit integers

i got this answer too in my mind as the first place can have 6 and the second can be 6 numbers to choose from .... 6 * 6 * 5 * 4 *3 /4 = 540 also
but i don't know why the program gives me 588 permutation — D0mBas3, Jun 04 '20 at 13:47

How many five digit integers can we make chosen our digits from 5,8,8, 2, 6, 6, 0?

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