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Why does it's stops at (n-3)? shouldn't it continue until (n-x)=1?

Why does it's stops at (n-3)? shouldn't it continue until (n-x)=1?-example-1
User Sungkwangsong
by
2.9k points

1 Answer

17 votes
17 votes

Solution

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given combination expression


^nC_4

STEP 2:Write the formula for combination


^nC_r=(n!)/((n-r)!r!)

STEP 3: Substitute the values


^nC_4=(n!)/((n-4)!4!)

Rewrite n!


n!=n(n-1)(n-2)(n-3)(n-4)!

We stop at (n-4)! so that it can be used to cancel out the (n-4)! which is the denominator of the combination expression. Therefore, we have:


\begin{gathered} (n(n-1)(n-2)(n-3)(n-4)!)/((n-4)!4!) \\ \\ (n-4)!\text{ cancels out each other to have:} \\ (n(n-1)(n-2)(n-3))/(4!) \\ 4!=4*3*2*1 \\ We\text{ have:} \\ (n(n-1)(n-2)(n-3))/(4*3*2*1)=(n(n-1)(n-2)(n-3))/(24) \end{gathered}

Hence, the reason for having the expression in the image question.

Why does it's stops at (n-3)? shouldn't it continue until (n-x)=1?-example-1
User Cleven
by
2.6k points
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