1. ## Distribution problem

I'm lost on these two.

7. How many different 7-digit number begin with an odd digit, end with an even digit, and do not repeat any digit for the remaining three digits?

11. How many ways are there to distribute 30 identical pens among four students so that the first student receives at least 3 pens, the second at least two pens, and the fourth receives EXACTLY four pens.

The "exactly" part threw me off in this one.

Help? thank you!

2. Originally Posted by domain07
I'm lost on these two.

7. How many different 7-digit number begin with an odd digit, end with an even digit, and do not repeat any digit for the remaining three digits?

11. How many ways are there to distribute 30 identical pens among four students so that the first student receives at least 3 pens, the second at least two pens, and the fourth receives EXACTLY four pens.

The "exactly" part threw me off in this one.

Help? thank you!
7)

C(n,k) = n choose k

There are C(5,1) ways to choose the first digit, C(5,1) ways to choose the end digit, obviously no overlap. If even and odd were switched then of course we'd have C(4,1) ways to choose the first digit because 0 would not be an option.

There are 5 digits remaining to be placed, and 8 choices. This makes C(8, 5).

There are 5! ways to permute the central digits.

So I think the overall answer should be C(5,1)*C(5,1)*C(8,5)*5!

11)

The pens are identical, so immediately we can take the last student out of the picture. The equivalent problem states:

How many ways are there to distribute 26 identical pens among three students so that the first student receives at least 3 pens and the second at least two pens?

Maybe you can take it from there.

EDIT: I just noticed something.

"7. How many different 7-digit number begin with an odd digit, end with an even digit, and do not repeat any digit for the remaining three digits?"

Shouldn't that be five instead of three?

EDIT 2: It was a bit silly of me to write C(5,1) above, when I could have just written 5.

3. Originally Posted by undefined
7)

C(n,k) = n choose k

There are C(5,1) ways to choose the first digit, C(5,1) ways to choose the end digit, obviously no overlap. If even and odd were switched then of course we'd have C(4,1) ways to choose the first digit because 0 would not be an option.

There are 5 digits remaining to be placed, and 8 choices. This makes C(8, 5).

There are 5! ways to permute the central digits.

So I think the overall answer should be C(5,1)*C(5,1)*C(8,5)*5!

11)

The pens are identical, so immediately we can take the last student out of the picture. The equivalent problem states:

How many ways are there to distribute 26 identical pens among three students so that the first student receives at least 3 pens and the second at least two pens?

Maybe you can take it from there.

EDIT: I just noticed something.

"7. How many different 7-digit number begin with an odd digit, end with an even digit, and do not repeat any digit for the remaining three digits?"

Shouldn't that be five instead of three?

EDIT 2: It was a bit silly of me to write C(5,1) above, when I could have just written 5.
An edit from my self, it is supposed to be 5 digit number, I made a mistake there. But I think I can figure it out for a five digit based on your help. Thank you! or else if you have the time, can you explain it to me with a 5 digit?

4. Originally Posted by domain07
An edit from my self, it is supposed to be 5 digit number, I made a mistake there. But I think I can figure it out for a five digit based on your help. Thank you! or else if you have the time, can you explain it to me with a 5 digit?
I get 5*5*C(8,3)*3!

Make sense?

Edit: I just learned about P(n, k) which gives the number of ordered k-subsets of {1,2,...,n}. This is in fact the same as C(n, k) * k! which I got above. But if you wanted to you could express as such:

5*5*P(8,3)

Saves time