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An algebra class has 21 students who all participate in a welcome activity introducing themselves to each other in pairs. One student wants to know how many introductions were made.

Fill in the values of [tex][tex]$n$[/tex][/tex] and [tex][tex]$r$[/tex][/tex] to complete the formula for finding the number of combinations of introductions.

[tex] \binom{n}{r} = \frac{n!}{r!(n-r)!} [/tex]


Sagot :

To find the number of introductions made when each student in a class introduces themselves to every other student in pairs, we use the combinatorial formula for combinations.

Given:
- [tex]\(n\)[/tex] is the total number of students in the class.
- [tex]\(r\)[/tex] is the size of each pair (which is 2 for pairs of students).

The formula for the number of combinations of choosing [tex]\(r\)[/tex] out of [tex]\(n\)[/tex] is given by:

[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]

We know the number of students, [tex]\(n\)[/tex], is 21, and we are choosing pairs, so [tex]\(r\)[/tex] is 2.

Now, substituting [tex]\(n = 21\)[/tex] and [tex]\(r = 2\)[/tex] into the formula:

[tex]\[ \binom{21}{2} = \frac{21!}{2!(21-2)!} = \frac{21!}{2! \cdot 19!} \][/tex]

Simplifying this, we get:

[tex]\[ \frac{21 \times 20 \times 19!}{2 \times 19!} = \frac{21 \times 20}{2} = \frac{420}{2} = 210 \][/tex]

So, the number of introductions made is 210.

Therefore, the values for [tex]\(n\)[/tex] and [tex]\(r\)[/tex] to complete the formula are:
[tex]\(n = 21\)[/tex] and [tex]\(r = 2\)[/tex].