Calc Help?

Question

5 P 2 = (200! / 198! 2!)

Answer 1

Your expression for the number of combinations is not correct. 5 p 2 should be 5!/3!2!. You expanded the factorials and then turned them back into factorials.

But here is the answer to the right side.
Since 200! is 200*199*198*197*196*... and 198! is 198*197*196*... then all of the factors from 198! will divide out from the 200!, leaving only 200*199 in the numerator and 2! (=2) in the denominator. The 2 divides into the 200 leaving 100*199.

So your answer is 19900. Didn't even need a calculator. This would be the number of combinations of 200 objects taken 2 at a time.

Back to the original question. It is solved the same way
5!/3!2! is 5*4*3*2*1/((3*2*1)*(2*1))
So you have 5*4/2 or 10 different combinations.

1&2
1&3
1&4
1&5
2&3
2&4
2&5
3&4
3&5
4&5

It works!