If the reading club has nine members, how many ways can the club choose five members to attend a special program if order does not matter?

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• David
Lv 7
1 month ago

Any 5 from 9 = 126 using your calculator

• 1 month ago

9C5 =

9!/(5!(9 - 5)!) =

126

• 1 month ago

It is case of finding combinations of 5 out of 9 members ( since order does not matter).  Formula used is -------

..   ...  .    ...n !

ⁿCᵣ = -------------------  .  In this case  n = 9  and  r = 5

...    .. r !  *  (n-r) !

.................    9!  ................9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

ie  ⁹C₅  = ---------------  =  -------------------------------------------  =  126

...       ...    5! * (9-5)!     .... 5 * 4 * 3 * 2 * 1  x  4 * 3 * 2 * 1

• 1 month ago

thats up to them who they choose

• 1 month ago

The answer is "9 choose 5"

To calculate that, start by writing 9/5 as a fraction:

9

--

5

Now multiply by the next smaller number on top and bottom. Repeat until you get to 1 on the bottom.

9 * 8 * 7 * 6 * 5

--------------------

5 * 4 * 3 * 2 * 1

If you are doing this by hand, you can cancel everything on the bottom. For example, you can cancel 5 and 5. You can also cancel 8 and 4 * 2. Finally you can partially cancel 9 and 3.

3 * 1 * 7 * 6 * 1

--------------------

1 * 1 * 1 * 1 * 1

Get rid of all the ones.

3 * 7 * 6

= 126 ways

As a double-check, go to Google and type "9 choose 5"

• alex
Lv 7
1 month ago