Permutations of the number of players on a baseball team

[six789]

i cannot get the question since i don't 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

 

[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.

 

[six789]

but it sounds as though he's asking how many different ways you can arrange 8 people in order.

shouldn't be 9 players they are asking?

 

[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.

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

 

[six789]

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

 

[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

 

[six789]

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

 

[ceptimus]

shouldn't be 9 players they are asking?

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.

 

[six789]

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

 

[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

 

[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?

 

[six789]

new question..

 

[six789]

how many of our arrangements would logically be odd or even?

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

 

[Gale]

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

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

 

[six789]

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

 

[cepheid]

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

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.

 

[cepheid]

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?

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?"

 

[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?

 

[cepheid]

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

then half would be odd, half are even, eh?

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