Question

I need help Question

Answer 1

The permutation and the combination would be ^{15}P_4= 32760 and ^{15}C_4 = 1365.

What is the formula used for permutation and combination?

To calculate permutation

$^nP_k = \dfrac{n!}{(n-k)!}$

Given; n = 15, k = r = 4

$^{15}P_4 = \dfrac{15!}{(15-4)!}\\\\^{15}P_4 = \dfrac{15!}{(11)!}\\\\^{15}P_4 = 15 \times 14 \times 13 \times 12 \\\\\\^{15}P_4 = 32760$

Therefore, ^{15}P_4 = 32760

For combination

$^mC_k = \dfrac{m!}{k! \times (m-k)!}$

$^{15}C_4 = \dfrac{15!}{4! \times (15-4)!}\\\\^{15}C_4 = \dfrac{15!}{4! \times (11)!}\\\\\\^{15}C_4 = \dfrac{32760}{4!}\\\\^{15}C_4 = 1365$

Learn more about permutations here:

brainly.com/question/13443004

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Answer 2

Question 5

$_{8}\text{P}_{2}=\frac{8!}{(8-6)!}=\frac{8!}{6!}=\frac{8 \cdot 7 \cdot 6!}{6!}=\boxed{56}$

Question 6

$_{15}\text{P}_{4}=\frac{15!}{(15-4)!}=\frac{15!}{11!}=\frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11!}{11!}=\boxed{32760}$