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Sagot :
To determine the number of ways the dance instructor can choose four of his ten students to be on stage for a performance, we use the concept of combinations, where the order does not matter. The number of combinations can be calculated using the binomial coefficient formula, commonly written as [tex]\( \binom{n}{r} \)[/tex].
Here, [tex]\( n \)[/tex] represents the total number of students, and [tex]\( r \)[/tex] represents the number of students to be chosen. In this problem, [tex]\( n = 10 \)[/tex] and [tex]\( r = 4 \)[/tex].
The formula for combinations is given by:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
Plugging in the values [tex]\( n = 10 \)[/tex] and [tex]\( r = 4 \)[/tex]:
[tex]\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} \][/tex]
[tex]\[ \binom{10}{4} = \frac{10!}{4! \cdot 6!} \][/tex]
We can simplify this by calculating the factorials:
[tex]\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \][/tex]
Next, simplify the expression by canceling out the [tex]\( 6! \)[/tex] term in both the numerator and the denominator:
[tex]\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4! \cdot 6!} \][/tex]
[tex]\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4!} \][/tex]
Now, calculate [tex]\( 10 \times 9 \times 8 \times 7 \)[/tex]:
[tex]\[ 10 \times 9 = 90 \][/tex]
[tex]\[ 90 \times 8 = 720 \][/tex]
[tex]\[ 720 \times 7 = 5040 \][/tex]
So, we have:
[tex]\[ \binom{10}{4} = \frac{5040}{24} \][/tex]
Dividing the numerator by the denominator:
[tex]\[ \frac{5040}{24} = 210 \][/tex]
Therefore, the number of different ways the instructor can choose four students out of ten is:
[tex]\[ \boxed{210} \][/tex]
Here, [tex]\( n \)[/tex] represents the total number of students, and [tex]\( r \)[/tex] represents the number of students to be chosen. In this problem, [tex]\( n = 10 \)[/tex] and [tex]\( r = 4 \)[/tex].
The formula for combinations is given by:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
Plugging in the values [tex]\( n = 10 \)[/tex] and [tex]\( r = 4 \)[/tex]:
[tex]\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} \][/tex]
[tex]\[ \binom{10}{4} = \frac{10!}{4! \cdot 6!} \][/tex]
We can simplify this by calculating the factorials:
[tex]\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \][/tex]
Next, simplify the expression by canceling out the [tex]\( 6! \)[/tex] term in both the numerator and the denominator:
[tex]\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4! \cdot 6!} \][/tex]
[tex]\[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4!} \][/tex]
Now, calculate [tex]\( 10 \times 9 \times 8 \times 7 \)[/tex]:
[tex]\[ 10 \times 9 = 90 \][/tex]
[tex]\[ 90 \times 8 = 720 \][/tex]
[tex]\[ 720 \times 7 = 5040 \][/tex]
So, we have:
[tex]\[ \binom{10}{4} = \frac{5040}{24} \][/tex]
Dividing the numerator by the denominator:
[tex]\[ \frac{5040}{24} = 210 \][/tex]
Therefore, the number of different ways the instructor can choose four students out of ten is:
[tex]\[ \boxed{210} \][/tex]
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