Question

1. how many combinations without repetition are possible if n = 7 and r = 2?

28
49
14
21

Explanation:

Step1: Recall combination formula

The formula for combinations without repetition is \( C(n, r)=\frac{n!}{r!(n - r)!} \), where \( n!=n\times(n - 1)\times\cdots\times1 \), \( 0!=1 \).

Step2: Substitute \( n = 7 \), \( r = 2 \)

First, calculate factorials: \( 7! = 7\times6\times5\times4\times3\times2\times1 \), \( 2! = 2\times1 \), \( (7 - 2)!=5!=5\times4\times3\times2\times1 \).
Substitute into the formula: \( C(7, 2)=\frac{7!}{2!(7 - 2)!}=\frac{7!}{2!5!} \).
Simplify: \( \frac{7\times6\times5!}{2\times1\times5!}=\frac{7\times6}{2\times1} \).

Step3: Calculate the result

\( \frac{7\times6}{2\times1}=\frac{42}{2}=21 \).

Answer:

21