Probability and 20 girls

[bhumika.k.shah]

Twenty girls qualify for the girls' Baseball team. How many teams of 9 girls each
may be formed? For each team of 9 girls, in how many different ways can the 9
starting positions be assigned?

I seriously am blank as to how to start with the problem!

[thephoenix]

IS IT 20C9*9P1

[ajith]

Twenty girls qualify for the girls' Baseball team. How many teams of 9 girls each
may be formed? For each team of 9 girls, in how many different ways can the 9
starting positions be assigned?

I seriously am blank as to how to start with the problem!

1. 20C9 ways

Selecting 9 gals from 20

2. It is 9! ways (I have no clue about baseball rules and positions)

[bhumika.k.shah]

Is your answer either of the two ?

OR

better to ask whats ur final answer

Twenty girls qualify for the girls' Baseball team. How many teams of 9 girls each
may be formed? For each team of 9 girls, in how many different ways can the 9
starting positions be assigned?

I seriously am blank as to how to start with the problem!

1. 20C9 ways

Selecting 9 gals from 20

2. It is 9! ways (I have no clue about baseball rules and positions)

[ajith]

Is your answer either of the two ?

OR

better to ask whats ur final answer

Twenty girls qualify for the girls' Baseball team. How many teams of 9 girls each
may be formed? For each team of 9 girls, in how many different ways can the 9
starting positions be assigned?

I seriously am blank as to how to start with the problem!

1. 20C9 ways

Selecting 9 gals from 20

2. It is 9! ways (I have no clue about baseball rules and positions)

If you observe the Original Question, It has two Questions. If you wanna combine questions, you have to multiply my answers

[Stuart@KaplanGMAT]

Twenty girls qualify for the girls' Baseball team. How many teams of 9 girls each
may be formed? For each team of 9 girls, in how many different ways can the 9
starting positions be assigned?

I seriously am blank as to how to start with the problem!

Let's look at this as 3 different questions:

1) how many different groups of 9 can be formed out of 20 people?

How to solve: use the combinations formula.

If you are choosing k items out of a group of n objects, there are n!/k!(n-k)! possible selections.

20C9 = 20!/9!11! (way too much math to appear on the actual GMAT other than in this form)

2) how many different ways can 9 people be arranged?

How to solve: use the simple permutations formula.

If you are arranging all the n members of a group, there are n! possible arrangements.

9! (too big a number to appear on the actual GMAT other than in this form)

3) how many different ways can we arrange 9 people out of a group of 20 people?

How to solve: use the permutations formula.

If you are arranging k members out of a group of n objects, there are n!/(n-k)! possible arrangements.

20P9 = 20!/(20-9)! = 20!/11! (way too much math to appear on the GMAT other than in this form)

* * *

For (3), we could also multiply the first two results:

20C9 * 9! = 20!/9!11! * 9! = 20!/11!

(since nPk = n!/(n-k)! and nCk = n!/k!(n-k)!, if we multiple nCk by k! we end up with the permutations formula).

[bhumika.k.shah]

Hi Stuart,
thanks for the brilliant explanation!
I just got stuck here, could you please explain me

(since nPk = n!/(n-k)! and nCk = n!/k!(n-k)!, if we multiple nCk by k! we end up with the permutations formula).[/quote]

[Stuart@KaplanGMAT]

Hi Stuart,
thanks for the brilliant explanation!
I just got stuck here, could you please explain me

(since nPk = n!/(n-k)! and nCk = n!/k!(n-k)!, if we multiple nCk by k! we end up with the permutations formula).

Hi,

that was just to demonstrate why both methods work.

When we look at combinations, order doesn't matter, so we factor out k! in order to eliminate duplicates.

When we look at permutations, order does matter, so there's no need to factor out k!.

The only difference between the two formulae is the k! in the denominator of the combinations formula. So, if we multiply the combinations formula by k!, we end up with the permutations formula.