How many arrangements of the four wins and three losses?

G

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A field hockey team played seven games and won four of them. There were no ties.

a) How many arrangements of the four wins and three losses are possible?

I thought it was like: 7C4 * 7C3 (35*35), since it says AND in between the question, but the answer in the back of the book is only 35..why


thanks
 
Try this:

\(\displaystyle \L\\\frac{7!}{4!3!}\)

You have 7 games total with 4 identical choices and 3 identical 'other' choices.


Think of it as "how many ways can you arrange the letters AAAABBB".
 
okay I get that, but why does order count in this question?
 
Suppose that the two teams are A & B.
Then the string ABAABBA resents one way for team A to win four games.
Any rearrangement of that string, four A’s & three B’s, represents four wins for A.
The number of ways that can be done is \(\displaystyle \frac{{7!}}{{\left( {4!} \right)\left( {3!} \right)}}\).
 
Hello, Anna

Another approach . . .

A field hockey team played seven games and won four of them. There were no ties.

a) How many arrangements of the four wins and three losses are possible?
Click to expand...

The team plays seven games: ._ _ _ _ _ _ _

Which four games did they win?
. . There are \(\displaystyle C(7,4)\,=\,35\) choices.

For each of these choices, say, W _ _ W W _ W
. . . . . . . . . . . . . . . . . . . . - - . . ↑ ↑ . . . . .↑
. . the 3 losses go in the empty slots.

 
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