Question

14. Simplify 3! (1 point)
2
5
3
6

15. Simplify 8P3. (1 point)
42
336
40,432
56

16. Simplify 15C3. (1 point)
182
455
2,730
910

Answer 1

Hey :)

Welcome to OpenStudy :)

Answer 2

14. Simplify 3! (1 point)
\[3 \times 2 \times 1 \]
--can you do this?

Answer 3

6

Answer 4

15. Simplify 8P3. (1 point)
--this is a permutation
\[P(n,r) = nPr = \frac{ n! }{ (n-r)! }\]
--so 8 is n & the 3 is the r
--plug them into the formula:
\[\frac{ 8! }{ (8-3)! }\]

Answer 5

yes #14.) is 6

Answer 6

thanks for 14

Answer 7

16. Simplify 15C3. (1 point)
-- this is a combination
\[C(n,r) = nCr = \frac{ n! }{ r! (n -r)! }\]
-- so the 15 is the n & the 3 is the r
--plug them into the formula:
\[\frac{ n! }{ 3! (15 -3)! }\]

Answer 8

i dont get 15 and 16

Answer 9

oops * forgot the n
\[\frac{ 15! }{ 3! (15 -3)! }\]

Answer 10

15 would be 336 right?

Answer 11

15.
\[ \frac{ 8! }{ (8-3)! }\]
\[ \frac{ 8! }{ 5! }\]
\[ \frac{ 8\times7\times6\times5\times4\times3\times2\times1 }{ 5\times4\times3\times2\times1 }\]
--cancel out from the 5 and after.
--so ur left with \[8\times7\times6\]
--what does this equal?

Answer 12

so 15.) is 336 your right :)

Answer 13

wb 16

Answer 14

have you tried it?

Answer 15

yes but its complicated

Answer 16

\[ \frac{ 15! }{ 3! (15 -3)! }\]
\[ \frac{ 15! }{ 3! (12)! }\]
\[ \frac{ 15\times14\times13\times \times12! }{ 3\times2\times1 \times 12! }\]
--the 12! cancel so what are you left with?

Answer 17

--sorry for the double x x
--its only one x

Answer 18

answer is 455

Answer 19

yes it is 455 :)

Answer 20

help with this please