1. ## Permutations

How many 7-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6?

I understand this part:

There are four cases:

1 is the first digit, 6 is the last digit << Case 1

2 is the first digit, 6 is the last digit<< Case 2

2 is the first digit, 2 is the last digit<< Case 3

1 is the first digit, 2 is the last digit <<Case 4

But. This is the part I don't understand...
Why is it that for case 1, the answer is
5! / (2!2!) ?

How do we get the 5 and why do we use 5 and not 7? I understand the denominator, since there are duplicates of 2 AND 5, but why 5 in the numerator and not 7?

For case 2 to 4, the answer is

5! / 2!

Why is it 5 and not 7 again? and how come this time there is only one duplicate taken into consideration in the denominator? What happens to the other one?

Please explain this as clearly as possible. Thank you for your help.

2. Hello, skeske1234!

How many 7-digit even numbers less than 3,000,000 can be formed
using the following digits: 1, 2, 2, 3, 5, 5, 6 ?

I understand this part:

There are four cases:

1 is the first digit, 6 is the last digit << Case 1

2 is the first digit, 6 is the last digit << Case 2

2 is the first digit, 2 is the last digit << Case 3

1 is the first digit, 2 is the last digit << Case 4

But this is the part I don't understand . . .
Why is it that for case 1, the answer is: $\frac{5!}{2!\,2!}$ ?

How do we get the 5 and why do we use 5 and not 7?
I understand the denominator, since there are duplicates of 2 and 5,
but why 5 in the numerator and not 7?

In Case 1, the first digit is 1, the last is 6: .1 _ _ _ _ _ 6

And we must place the other five digits {2, 2, 3, 5, 5} in those spaces.
. . And there are: . $\frac{5!}{2!\,2!}$ ways.

In Case 3, the first digit is 2, the last is 2: .2 _ _ _ _ _ 2

And we place the other five digits {1, 3, 5, 5, 6} in those spaces.
. . And there are: . $\frac{5!}{2!}$ ways.

Get the idea?