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## #1 2006-01-28 13:00:12

silvercity87
Member
Registered: 2006-01-28
Posts: 5

### Permutation

The question goes:
How many 6-digit odd numbers less than 200 000 can be formed using the digits 1,1,2,2,3 and 5?
How do I solve this kind of question? can anybody help me please?!!!:

Last edited by silvercity87 (2006-01-28 13:00:52)

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## #2 2006-01-28 13:09:45

God
Member
Registered: 2005-08-25
Posts: 59

### Re: Permutation

you have 6 digits.
So, there are 6 choices for the hundred thousands digit, 5 choices (all the numbers except whatyou already used) for the ten thousands digit, 4 choices (all the numbers except the two you used) for the thousands digit, 3 for the hundreds, 2 for the tens, and 1 (whatever is left over) for the one's digit.

This makes 6*5*4*3*2*1 = 720

However, since there are two 1's and two 5's, we need to divide by 2*1 = 2 twice
(if you have two one's, it doesn't matter which 1 goes in which place. Since there are two ways to order two things, we divide by two)

720/2/2 =

permutations

Last edited by God (2006-01-28 13:10:38)

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## #3 2006-01-28 13:27:08

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Permutation

The basic method is right, but God forgot to consider that the must be less than 200 000.

For this to work, the first number has to be 1. After that, you use a similar method to God's to find the number of ways you can arrange the other numbers.

5*4*3*2*1 = 120, but there are 2 2's, so divide by 2 to get

.

Why did the vector cross the road?
It wanted to be normal.

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## #4 2006-01-28 13:28:13

silvercity87
Member
Registered: 2006-01-28
Posts: 5

### Re: Permutation

But the number has to be odd as well :\$

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## #5 2006-01-28 14:58:23

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Permutation

Oh dear, we're all making mistakes now!

OK then, the same thing applies, but there are now only 3 choices for the last digit (1, 3 and 5).

Now we have 3*4*3*2*1 = 72, but we have to divide by 2 again because of the 2 2's, so we get a final answer of

.

There's probably something wrong with that too.

Why did the vector cross the road?
It wanted to be normal.

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## #6 2006-01-28 16:33:02

silvercity87
Member
Registered: 2006-01-28
Posts: 5

### Re: Permutation

So in a similar question that goes
"How many 7 digit even numbers less than 3 000 000 can be formed using all the digits 1,2,2,3,5,5,6?", the answer would be

3*5*4*3*2*1*2 divide by 2*2=180 right?

I'm confused
Thank you for the answers though!

Last edited by silvercity87 (2006-01-28 16:39:23)

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## #7 2006-01-28 22:19:28

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Permutation

Hmm. This one's harder, because the constraints don't fully overlap.

You've gone wrong where you said that there were 3 possible choices for the first digit (1, 2 and 2) and that there were two possible for the last one (2 and 6). If you took the first digit to be 1, then the other 2 could also be a possibility for the last digit.

That would make the answer slightly more than what you got, which is what it is, so presumably you've done everything else correctly. It certainly looks right.

Why did the vector cross the road?
It wanted to be normal.

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