Question

O PROBABILITY AND STATISTICSComputing permutations and combinationsEvaluate 10C6 and 10510% = 0P₁ = 010 5I need help with this math problem.

Answer 1

Explanation

First part

The formula for combinations is:

$_nC_r=\frac{n!}{(n-r)!r!}$

On the other hand, n! is the product of all positive integers less than or equal to n. For example:

$3!=3*2*1=6$

Then, we have:

$\begin{gathered} _nC_r=\frac{n!}{(n-r)!r!} \\ _{10}C_6=\frac{10!}{(10-6)!6!} \\ _{10}C_6=\frac{10!}{4!6!} \\ _{10}C_6=\frac{10*9*8*7*6*5*4*3*2*1}{(4*3*2*1)(6*5*4*3*2*1)} \\ _{10}C_6=\frac{3,628,800}{24*720} \\ _{10}C_6=\frac{3,628,800}{17,280} \\ _{10}C_6=210 \end{gathered}$

Second part

The formula for permutations is:

$_nP_r=\frac{n!}{(n-r)!}$

Then, we have:

$\begin{gathered} _{n}P_{r}=\frac{n!}{(n-r)!} \\ _{10}P_5=\frac{10!}{(10-5)!} \\ _{10}P_5=\frac{10!}{5!} \\ _{10}P_5=\frac{10*9*8*7*6*5*4*3*2*1}{5*4*3*2*1} \\ _{10}P_5=\frac{3,628,800}{120} \\ _{10}P_5=30,240 \end{gathered}$Answer$\begin{gathered} _{10}C_{6}=210 \\ _{10}P_{5}=30,240 \end{gathered}$