# Permutations of the number of players on a baseball team

1. Nov 5, 2005

### six789

i cannot get the question since i dont know how to play baseball...
this is the question...
The manager of a baseball team has picked the nine players for the starting lineup. In how many ways can he set the batting order so that the pitcher bats last?

im just guessing, is this correct?
8*7*7 = 392

therefore, the pitcher can set the batting order in 392 ways

Last edited: Nov 5, 2005
2. Nov 6, 2005

### ceptimus

I don't know baseball either, but it sounds as though he's asking how many different ways you can arrange 8 people in order.

Any of the 8 can go first. For each of those 8 alternatives, you can choose from the remaining 7 to go second. So there are 8 x 7 ways of choosing the first two.

And for each of those 56 alternatives you can choose from the 6 remaining players which one bats third and so on...

8 x 7 x 6 x 5 x 4 x 3 x 2

There is a special button on scientific calculators for this sort of calculation. It's written 8! and the exclamation point is pronounced factorial.

3. Nov 6, 2005

### six789

shouldn't be 9 players they are asking?

4. Nov 6, 2005

### Jameson

This is how I would go about it. Since the pitcher has to be last, you only need to worry about the positioning of the other 8 people. And since order matters for this positioning, you are going to use permutations.

$$nPr=\frac{n!}{(n-r)!}$$

5. Nov 6, 2005

### six789

ohhh i see, why is the formula different from what i learned, coz it is nPr = n!/(p!q!r!)? im soo confused now on the formulas

6. Nov 6, 2005

### Jameson

I don't know where you learned that formula, and it doesn't make sense. P isn't a number, it just means "permutation".

Look at this mathworld site or Google "permutation" to check me.

http://mathworld.wolfram.com/Permutation.html

7. Nov 6, 2005

### six789

ill check the site... thanks for the addiional info

8. Nov 6, 2005

### ceptimus

I don't think so. If the pitcher has to bat last, then you don't have to worry about him - only the other 8 can be rearranged in order.

9. Nov 6, 2005

### six789

ok, thanks for guiding me to the right way cetimus...

10. Nov 6, 2005

### six789

ceptimus, is my work on the right track?

is it 9!/5! or 9x8x7x6 for 4d? since the numbers runs from 0 to 9, but there is only 5 odd numbers, and same thing with 4e?

this is the question...
4. How many four digit numbers are there with the following retrictions?
d) the number is odd
e) the number is even

11. Nov 6, 2005

### Gale

is this a new question or is it still about baseball? cause it makes no sense if its still about baseball.........

so you're taking 10 digits, (0-9) and arranging them into 4 places right? this is the exact definition of a permutation. P(10,4). now, what makes a number odd or even? only the last didgit. how many of our arrangements would logically be odd or even?

12. Nov 6, 2005

### six789

new question..

13. Nov 6, 2005

### six789

for odd, it is 9!/5! and for even, it is 9!/4!

Last edited: Nov 6, 2005
14. Nov 6, 2005

### Gale

why? if we have a million chronological numbers, how many will be odd? half right? the other half are even!

15. Nov 6, 2005

### six789

so my answer for odd is right and for even it is wrong?

Last edited: Nov 6, 2005
16. Nov 6, 2005

### cepheid

Staff Emeritus
Read what she said. Half the numbers are even, and half are odd. So to know how many there are of each type, all you have to do is figure out how many four-digit numbers there are in total.

17. Nov 6, 2005

### cepheid

Staff Emeritus
Quick question, Gale. If you just choose any number from 0-9 for each of the four places, would that allow for permutations like: 0000, and 0145? Sure a number can have as many leading zeros as you like, but do those really count as "four-digit numbers?"

18. Nov 7, 2005

### Gale

ya, hmm... i spose you're right, actually, you can also use didgets twice. so, i guess if you only use 1-9 for the first place, then you can use all 10 in the other three? 9*10*10*10 would be all the possible numbers. and then half would be odd, half are even, eh?

19. Nov 7, 2005

### cepheid

Staff Emeritus
Yeah I agree, we're still permuting. Looks ok to me.

Picking up (so-called) Canadian habits, eh? From who, I wonder? We haven't even talked in a while.