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: ?

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 otherfivedigits {2, 2, 3, 5, 5} in those spaces.

. . And there are: . ways.

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

And we place the otherfivedigits {1, 3, 5, 5, 6} in those spaces.

. . And there are: . ways.

Get the idea?