Permutations.

**Big-K** · January 25th 2009, 09:14 PM

10C2
i know the answer is 45.
but for some reason, i can't get THAT answer. i keep getting something else....

anyone help me please?

thank you in advance!

**andreas** · January 25th 2009, 09:19 PM

Originally Posted by Big-K

10C2
i know the answer is 45.
but for some reason, i can't get THAT answer. i keep getting something else....

anyone help me please?

thank you in advance!

10 C 2= $\frac{10!}{8!2!}$ after dividing denomonator and numerator by 8! the following is obtained $\frac{(9*10)}{(1*2)}=9*5=45$

**Big-K** · January 25th 2009, 09:28 PM

But does this 10 C 2=

have anything to do with the 2nd step?
if it does, how?

thank you!

**andreas** · January 25th 2009, 09:44 PM

Originally Posted by Big-K

But does this 10 C 2=

have anything to do with the 2nd step?
if it does, how?

\frac{10*9*8*7*6*5*4*3*2*1}{8*7*6*5*4*3*2*1*2*1}

thank you!

well ,I assumed that you know this:

$10!=10*9*8*7*6*5*4*3*2*1$

$8!=8*7*6*5*4*3*2*1$

and $2!=2*1$

Substituting this gives:

$\frac{10!}{8!2!} = \frac{10*9*8*7*6*5*4*3*2*1}{8*7*6*5*4*3*2*1*2*1} = \frac{9*10}{2}=45$

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