I need help with the following math problem?

Question

Of the 12 people who show up, 2 are women. How many ways are there to choose 10 players to take the field if at least one of these players must be women?

Please show work or atleast a setup. I've made several failed attempts to solve this problem?

Answer 1

There are 10C9 ways to pick 9 men to be on the team
10C9 =10!/9!1! = 10
There are two ways to pick 1 of 2 women
Hence there are 20 ways to pick 9 men and one woman.

Answer 2

this may be a trick question not as a results of content fabric, yet as a results of assumed answer. there is not any lacking dollar. The area that every person messes up on is the equation. it extremely is right that $9 x 3= $27 and that the bell boy has $2, yet you may not upload ($27 + $2) jointly. Why? because of the fact, If the bell boy is given $5 returned, and he wallet $2, then there is in common terms $3 left. The bell boy then divides that $3 into 3 (for all of the adult adult males). which skill each and each guy gets $a million, not $3, which equals $3 money returned in total. So, the quantity lacking is $2, which the bell boy already has. Plus, the value of the hotel is not $30 yet $25. subsequently, the breakdown is as follows: the value of the hotel is $25 ($30-$5). The bell boy has $2. each and all of the adult adult males have $a million = $3 in total. 25 + 2 + a million+ a million + a million = 30. not something lacking, nevertheless $30.

Answer 3

There can be either 1 woman or 2 womwn in the group.

If there are 9 men and 1 woman:
We choose 9 men out of 10: (10!)/(9!*1!) = 10 ways to do this
We choose 1 woman out of 2: (2!)/(1!*1!) = 2 ways to do this
So there are 10*2 = 20 ways

Or there can be 8 men and 2 women:
We choose 8 men out of 10: (10!)/(8!*2!) = 45
We choose 2 woman out of 2: (2!)/(2!*1!) = 1
So there are 45*1=45 ways to do this

So there are 20+45=65 ways total

Answer 4

Here we go...

Combinations...

First lets start with just one of the women.

We get 10!/(9!*1!)=10. BUT we have two women, so we get 20...

Ok, now for both women. 10!/(8!*2!) which gives us 45

So 20+45 gives 65 possible combinations


Make sure not to confuse this with permutations. Permutations would mean there was an order to the players.

ex. 1324 and 1234 would be two different teams because of the order. That's not what you're looking for.

The link has the formula. I don't have time to derive it. Sorry :(

Answer 5

Obviously there are 12 people and if 1 woman, at least, has to take the field then u r left with nine players to chose from.
10*11=112 diff ways to organize the players.(dont forget to add girl)

Answer 6

At the root, the team must contain:

Woman A
or
Woman B
or
Woman A + Woman B

See if you can figure out the combinations of the other people who can be used with one of the above combinations.

Answer 7

2 ways

Answer 8

12*10 or 10*10 If you multiply the total number of people by how many are going to the field you should get your answer.

I hope you get it!!!

Answer 9

Why don't you like...
Leave it blank, and ask the teacher about it tomarrow?
That's what I do, and it helps more than getting 20 different answers from different people!

Answer 10

That makes my head hurt.