To determine how many different pairs of students can be formed from a class of 21 students, we can use the formula for combinations. The formula for combinations is given as:
[tex]\[ C(n, r) = \frac{n!}{r! \cdot (n - r)!} \][/tex]
Here, [tex]\( n \)[/tex] represents the total number of students, and [tex]\( r \)[/tex] represents the number of students in each pair. In this scenario, [tex]\( n = 21 \)[/tex] and [tex]\( r = 2 \)[/tex].
So, we plug in the values into the formula:
[tex]\[ C(21, 2) = \frac{21!}{2! \cdot (21 - 2)!} \][/tex]
Calculating the values:
1. [tex]\( 21! \)[/tex] is the factorial of 21,
2. [tex]\( 2! \)[/tex] is the factorial of 2, which equals 2,
3. [tex]\( (21 - 2)! = 19! \)[/tex] is the factorial of 19.
The number of combinations will be:
[tex]\[ C(21, 2) = \frac{21!}{2! \cdot 19!} \][/tex]
This simplifies to:
[tex]\[ C(21, 2) = \frac{21 \times 20 \times 19!}{2 \times 19!} \][/tex]
Since [tex]\( 19! \)[/tex] appears in both the numerator and the denominator, it cancels out:
[tex]\[ C(21, 2) = \frac{21 \times 20}{2} = \frac{420}{2} = 210 \][/tex]
Therefore, the total number of unique pairs of introductions is [tex]\( 210 \)[/tex].
