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# combination and permutation?

I have two solutions for this problem.

1. (5C3)(4C3)(6!) = 28800 (solution from my old algebra textbook)2. 5*4*3*4*3*2*6! = 1036800 (my solution)

What is your solution?What is the right solution?

### 5 Answers

- nbsaleLv 62 months ago
The first solution is the correct one. In your 2nd expression, replace 6! with 6C3. See my comment to sepia's answer for more on why 6! is wrong there.

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- Anonymous2 months ago
What I see is (5c*3)*(4c*3)*6!,That is 5c*4c=20c squared*

15c*12c=180c ////20c squared*180c *61//3600c cubed*61=219600c cubed

That is just a guess.

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- KrishnamurthyLv 72 months ago
How many 6-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8,

and 9 if each 6-digit number consists of 3 odd digits and 3 even digits?

Repetition of digits is not allowed.

123456, 135246, ....

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- sepiaLv 72 months ago
I have two solutions for this problem.

1.

(5C3)(4C3)(6!) = 28,800 (solution from my old algebra textbook)

2.

5*4*3*4*3*2*6! = 1,036,800 (your correct solution)

What is your solution?

The answer is 1,036,800.

What is the right solution?

5*4*3*4*3*2*6! = 1,036,800

- nbsaleLv 62 months agoReport
5*4*3*4*3*2*6! = 1,036,800 is wrong. 5*4*3*4*3*2 already accounts for the permutations of the odd and even numbers. You have 5 choices for the first odd, 4 for the 2nd, 3 for the 3rd. Now you need to multiply by 6C3, the number of choices for where to put the 3 selected. you get 1440*20 = 28800

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- 2 months ago
Order matters. 123456 is not the same as 654321.

You have 4 even numbers and 5 odd numbers

5C3 * 4C3 * 6! =>

(5! / (3! * (5 - 3)!)) * (4! / (3! * (4 - 3)!)) * 6! =>

(120 / (6 * 2)) * (24 / (6 * 1)) * 720 =>

10 * 4 * 720 =>

28800

Your answer is wrong, because you are using permutations, not combinations. I know that this seems contradictory to what I said before, but it isn't.

While 123456 and 654321 may be different numbers, they use the same combinations of digits, which is why we use combinations, not permutations to describe them.

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