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Mathematics 8 Online
OpenStudy (anonymous):

I have No clue how to do this problem can someone help please . Evaluate 9C5 126 24 3024 120

OpenStudy (amistre64):

9.8.7.6.5 -------- 5.4.3.2.1

OpenStudy (amistre64):

mine just cuts off the 4! and leaves the rest

OpenStudy (anonymous):

so take 9*8*7*6*5= 15120 and 5*4*3*2*1= 120 so the answer would be 120?

OpenStudy (anonymous):

Like amistre64 said, he left out the 4!, so divide your answer by 4!.

OpenStudy (anonymous):

3780 is answer

OpenStudy (zarkon):

\[\frac{9\times8\times7\times6}{4\times3\times2\times1}=126\]

OpenStudy (anonymous):

Ohhh wait that wouldn't give you an answer within the choices... oops

OpenStudy (anonymous):

Zarkon's looks correct

OpenStudy (anonymous):

\[nCr = \frac{n\!!}{r\!!(n-r)\!!}\] \[\implies 9C5 = \frac{9\!!}{5\!!(9-5)\!!} = \frac{9\!!}{5\!! 4\!!} = \frac{362880}{120 \cdot 24} = 126\]

OpenStudy (phi):

Although typically you would not multiply everything out until after canceling numbers in the top and bottom.

OpenStudy (anonymous):

\[9 C 5=\frac{9\times 8\times 7\times 6\times 5}{4\times 3\times 2}=9\times 2\times 7=126\]

OpenStudy (anonymous):

Phi has a point, it's a lot easier (and neater) to evaluate when you cancel out a few numbers.

OpenStudy (anonymous):

cancel first multiply last because "the number of ways to choose k from a set of n" is always in integer

OpenStudy (anonymous):

actually there is a typo in the middle part of my answer

OpenStudy (anonymous):

5 shouldn't be there.

OpenStudy (anonymous):

thank you for showing me how to do this problem :)

OpenStudy (amistre64):

9.8.7.6.5 -------- 5.4.3.2.1 9.8.7.6 -------- 4.3.2 9.2.7.6 -------- 3.2 9.2.7.2 -------- 2 9.2.7 = 63.2 = 126

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