Question

for the given values of n and r, evaluate (a) (\frac{n!}{(n - r)!}) and (b) (\frac{n!}{r!(n - r)!}) for (n = 11) and (r = 4).

(a) when (n = 11) and (r = 4), (\frac{n!}{(n - r)!} = 7920). (simplify your answer. type an integer or a simplified fraction.)

(b) when (n = 11) and (r = 4), (\frac{n!}{r!(n - r)!} = square). (simplify your answer. type an integer or a simplified fraction.)

Explanation:

Step1: Recall the formula for combinations

The formula given is \(\frac{n!}{r!(n - r)!}\), which is the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\). We know \(n = 11\) and \(r=4\), so first we can find \(n-r=11 - 4=7\).

Step2: Substitute the values into the formula

Substitute \(n = 11\), \(r = 4\) and \(n-r=7\) into the formula: \(\frac{11!}{4!(11 - 4)!}=\frac{11!}{4!×7!}\).
We know that \(n!=n\times(n - 1)\times\cdots\times1\), so \(11! = 11\times10\times9\times8\times7!\) and \(4! = 4\times3\times2\times1\).
Substitute these into the fraction: \(\frac{11\times10\times9\times8\times7!}{4\times3\times2\times1\times7!}\).

Step3: Simplify the fraction

The \(7!\) terms in the numerator and denominator cancel out. Then we calculate the remaining part: \(\frac{11\times10\times9\times8}{4\times3\times2\times1}\).
Calculate the numerator: \(11\times10\times9\times8=7920\).
Calculate the denominator: \(4\times3\times2\times1 = 24\).
Then \(\frac{7920}{24}=330\).

Answer:

330