***Hard Math question*** i tryed for 30mins to slove :(!?

Question

Ok heres the question ....

there are 5 players player1,player2,player3,p... captain can only pick 3 of those players..how many possible combinations are there for the captain? also how did you get your answer....if you include a mathematical formula your getting 5/5!

Answer 1

Here's how I did it:

can only use each number once in a set, order of numbers not important (123 same as 321).

So:

123, 124, 125,

134, 135,

145

234, 235,

245

345

Answer 2

you can pick the first person 5 ways, the second person 4 ways, and the third person 3 ways. So the possibilities would be 5x4x3 = 60 different possibilities.

Answer 3

Umm..i think it's ten?

If you only want the captain to choose each player only once..

I just made a chart  with the players and paired them of

Answer 4

Here's something that will explain everything in detail http://mathforum.org/dr.math/faq/faq.com...

It's 10 - you take [5!/(5-3)!]/3! = (120/2)/6 = 60/6 = 10

*note - ! means 'factorial' that's where you multiply the number by each numeral under it. 5! = 5x4x3x2x1

The above formula is for a combination. That is where order doesn't matter. Permutations are similar, but order matters.

***Hey, I really don't think it's 60 - I'm pretty sure it's 10. It would be 60 if the order was important (if it mattered which guy was picked first, second, and third)

Answer 5

n!

          [  c!(n-c)!  ] -1

To make it simple follow these procedures.

n=amount of players total.

!=factorial which multiplies the base number to the number desired in your case 5. this makes it so there is no repeat of posibilites of reciprocal ratios.

c=the number of player are to be chossen, which is 3 of those players and combine.

(-1)= the subtractor of any recprical combination.

The space between the top portion and bottom portion of the formula is division.

First plug the numbers in

Second do ALL factorials of both top and bottom portions

Third do the multiplication then factorial the last( ! )at the end of; c!(n-c)!  the factorial on the far right of that equation do that after multiplying,but do not munipulate it befor multipying;and when you factorial the first number n! and c!.

Last Subtract one if( nessessary) to eliminate all odds so that you dont have an extra number.

the rest is for you to do good luck. oh and divide after all that mess.

Answer 6

it's pretty much permutation. 5P3 = 5*4*3

which is 60