Question

how to find r in this equation using combination formula C(8,r)=28
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Answer 1

Answer:

r = 2

Step-by-step explanation:

We have the formula of $^nC_r = \frac{n!}{r! (n-r)!}$

Now, it is given that $^8C_r = \frac{8!}{r! (8-r)!} = 28$ ........ (1)

And we have to find the value of r which satisfy the above equation.

So, $r! (8-r)! = \frac{8!}{28} = \frac{40320}{28} = 1440$

Now, we have to use the trial method to find the value of r.

For r = 1, $1! (8-1)! = 7! = 5040 \neq 1440$

Hence, r can not be 1.

Now, put r = 2, $2! (8-2)! = 2 \times 6! = 1440$

Therefore, r = 2 (Answer)

Answer 2

To find r in this equation using combination formula C(8,r)=28 we used The value r=2

Explanation:

Permutations and combinations are the various ways where objects from a set may be selected without replacement to form subsets. Combination is a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate it we will use the formula nCr = n! / r! * (n - r)! where n is the total number of items, and r is the number of items being chosen at a time

By using combination C(8,r)=28 and using formula

$nCr = \frac{n!}{r!(n-1)}!}$

$28 = \frac{r!}{r!(8-r)!}$

$r!(8-r)!=\frac{8!}{7(4)}}$

$r!(8-r)!=\frac{8(7)(6!)}{(7)(4)}$

$r!(8-r)!=2(6!)$

So,

$r!=2 or (8-r)! = 6!$

$r=2 or {r \ \textless \ or = 8}$

The value $r=2$, because 2 is satisfied the given combonation

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