Combinations- how to show that 10C2=10C8

Mathematics

1 answer:

Black_prince [1.1K]3 years ago

7 0

By definition,

${}^nC_k=\dbinom nk=\dfrac{n!}{k!(n-k)!}$

We have

$\dbinom{10}2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10!}{(10-8)!8!}=\dbinom{10}8$

and so

${}^{10}C_8={}^{10}C_2$

You might be interested in

Solve the inequality algebraically. Write the solution in interval notation. |9x + 10| < 15

viva [34]

The given inequality is

$| 9x+ 10 |$

First we need to remove the absolute sign , and when we do so , we will get a compound inequality, that is

$-15< 9x+10 < 15$

To solve for x, first we need to get rid of 10. and for that, we have to do subtraction

$-15-10$

Now we need to get rid of 9, and for that, we do division

$\frac{-25}{9}< x< \frac{5}{9}$

So the required solution is

$( \frac{-25}{9} , \frac{5}{9})$

5 0

3 years ago

Find surface area and 136. 224 236 248 oops I forgot -.-|||

xxMikexx [17]

Answer:

B, 224in2

It may get tedious but just find the area of each face and write them down

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Raymond has taken a position as the navigator on a speed boat racing team. The team entered a race that starts at Daytona Beach,

erik [133]

Using the rules of cosines, the measure of the turn between the second and third legs of the race is given by

$\cos^{-1}\left( \frac{180^2+160^2-140^2}{2\times180\times160} \right)=\cos^{-1}\left( \frac{32,400+26,600-19,600}{57,600} \right) \\ \\ =\cos^{-1}\left( \frac{39,400}{57,600} \right)=\cos^{-1}(0.6840)\approx47^o$