freethinking, could you exlain how you did the following:

2. [5!/(2!*3!)]*5^5 - [4]*5^4 ( to remove numbers start with 0)

3. [5!/4!]*5^5 - [4]*5^4 ( to remove numbers start with 0)

number of ways

1)ODD1, ODD, ODD, ODD, ODD

2)ODD1, ODD, ODD, EVEN, EVEN

3)ODD1, EVEN, EVEN, EVEN, EVEN

1) can be accomplished in 5^5 ways

n1 = (5)^5

2) ODD1 can be chosen in 5 ways. Then we have to choose 2 ODD numbers and two EVEN numbers. They all can be choosen in 5 ways each and they can be placed in 4! ways.

n2 = (4!)*(5)^5 = 51000.

I know i commmitted an error here. I haven't subtracted the repititive combinations here. Could somebody help?

3) same thing can be done with this case.

somebody, please explain.

3) OD

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