Question

at a game show , there are 7 people (including you and your friend) in the front row. the hoat randomly chooses 3 people from the front row to be contestants. the order in witch they are chosen does not matter. how many ways can you and your frined be choosen

Answer 1

2x3:7 = 0,857

You and your friend have 0,857 chance to be choosen.

Answer 2

Answer:

5 ways

Step-by-step explanation:

There are 7 people (including you and your friend) in the front row.

The hoat randomly chooses 3 people from the front row to be contestants.

We are supposed to find how many ways can you and your friend be chosen.

Since you and your friend is fixed

So, total ways = $^2C_2$

Now one more is remaining for being selected .

Since two are selected out of 7 . So, 5 are remaining .

So, the last person will be selected from remaining 5  i.e. $^5C_1$

So, no. of ways  you and your friend can be chosen:

$^2C_2 \times^5C_1$

Formula : $^nC_r=\frac{n!}{r!(n-r)!}$

$\frac{2!}{2!(2-2)!} \times\frac{5!}{1!(5-1)!}$

$1 \times5$

$5$

Thus there are 5 ways in which you and your friend can be chosen