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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].
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].
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