Combinatorial question: permutation, binomial coefficient

[blob84]

How many numbers of 6 digits which have exatctly the digit 1 (2 times), digit 2 (2 times), without zero, are there?
The book post this solution: $$\frac{6!}{2!2!}*\binom{7}{2} + \frac{6!}{2!2!2!}*7= 4410$$,
but I'm trying to find an explanation for this result.

 

[mfb]

That is a weird way to calculate the number.

Here is what I would do:

There are $6 \choose 2$ ways to place the "1"s, and $4 \choose 2$ ways to place the "2"s afterwards. If you write both with factorials, a 4! cancels and you get $\frac{6!}{2! 2! 2!}$. The remaining two digits have 7 options each (3...9), therefore the total number of digits is...

Of course, you can split 7*7 in $2 {7\choose 2} + 7$, but where is the point? The first part corresponds to the number of numbers where the two remaining digits are different, but I don't see a reason to consider them separately.

 

[blob84]

It is $$\binom{6}{2}*\binom{4}{2}*7^2$$ by the general form of the product rule, where there are for any pairs of digits 3 cases, if I'm not wrong, thank you.
I suppose to get this weird formula it calculate permutation with repetition and it brokes the problem in two sets.

 

[ssd]

The 1st term corresponds to the case when the other two digits are different and the 2nd term corresponds to the case when another digit is repeated (three double digits).

 

[blue_raver22]

The combinatorial question provided is asking for the number of 6-digit numbers that have exactly two 1's and two 2's, without any zeros. This can be solved using the concept of permutations and binomial coefficients.

First, we can consider the arrangement of the digits 1 and 2 in the 6-digit number. This can be done using permutations, as we want to find the number of ways to arrange 2 1's and 2 2's in a 6-digit number. This can be calculated using the formula 6!/2!2!, which gives us 6*5*4*3*2*1 / (2*1*2*1) = 180.

Next, we need to consider the remaining digits in the 6-digit number. Since we do not want any zeros, we have 7 possible digits to choose from (1, 2, 3, 4, 5, 6, 7). We can choose any two of these digits to fill the remaining two spots in the 6-digit number. This can be calculated using binomial coefficients, specifically \binom{7}{2}, which gives us 7!/2!(7-2)! = 7*6 / (2*1) = 21.

Therefore, the total number of 6-digit numbers that meet the given criteria is 180*21 = 3780. However, we need to add the cases where the remaining two digits are both 1's or both 2's, as these numbers were already counted in the permutation calculation. This can be done by adding the number of 6-digit numbers with two 1's (6!/2!2!2!) and the number of 6-digit numbers with two 2's (6!/2!2!2!), which gives us 3780 + 630 + 630 = 4410.

In summary, the solution provided in the book is using the concept of permutations and binomial coefficients to calculate the total number of 6-digit numbers with exactly two 1's and two 2's, without any zeros.