Answer :
Sure, let's evaluate the expression [tex]\( C(10, 6) \)[/tex], which represents the number of combinations of choosing 6 items from a set of 10 items.
The formula for combinations is given by:
[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]
where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and [tex]\( ! \)[/tex] denotes factorial, which is the product of all positive integers up to that number.
For our specific case:
[tex]\[ n = 10 \quad \text{and} \quad k = 6 \][/tex]
So, substituting the values into the formula, we get:
[tex]\[ C(10, 6) = \frac{10!}{6! \cdot (10 - 6)!} = \frac{10!}{6! \cdot 4!} \][/tex]
Using the factorial values:
[tex]\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 \][/tex]
Substituting these into the formula:
[tex]\[ C(10, 6) = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times 4 \times 3 \times 2 \times 1} \][/tex]
Canceling the [tex]\( 6! \)[/tex] terms from the numerator and the denominator:
[tex]\[ C(10, 6) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \][/tex]
Calculating the remaining products and division:
[tex]\[ 10 \times 9 = 90 \][/tex]
[tex]\[ 90 \times 8 = 720 \][/tex]
[tex]\[ 720 \times 7 = 5040 \][/tex]
[tex]\[ 4 \times 3 = 12 \][/tex]
[tex]\[ 12 \times 2 = 24 \][/tex]
[tex]\[ 24 \times 1 = 24 \][/tex]
Finally, dividing the products:
[tex]\[ C(10, 6) = \frac{5040}{24} = 210 \][/tex]
Therefore, the value of [tex]\( C(10, 6) \)[/tex] is:
[tex]\[ 210 \][/tex]
This is the number of ways to choose 6 items from a set of 10 items.
The formula for combinations is given by:
[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]
where [tex]\( n \)[/tex] is the total number of items, [tex]\( k \)[/tex] is the number of items to choose, and [tex]\( ! \)[/tex] denotes factorial, which is the product of all positive integers up to that number.
For our specific case:
[tex]\[ n = 10 \quad \text{and} \quad k = 6 \][/tex]
So, substituting the values into the formula, we get:
[tex]\[ C(10, 6) = \frac{10!}{6! \cdot (10 - 6)!} = \frac{10!}{6! \cdot 4!} \][/tex]
Using the factorial values:
[tex]\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 \][/tex]
Substituting these into the formula:
[tex]\[ C(10, 6) = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times 4 \times 3 \times 2 \times 1} \][/tex]
Canceling the [tex]\( 6! \)[/tex] terms from the numerator and the denominator:
[tex]\[ C(10, 6) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \][/tex]
Calculating the remaining products and division:
[tex]\[ 10 \times 9 = 90 \][/tex]
[tex]\[ 90 \times 8 = 720 \][/tex]
[tex]\[ 720 \times 7 = 5040 \][/tex]
[tex]\[ 4 \times 3 = 12 \][/tex]
[tex]\[ 12 \times 2 = 24 \][/tex]
[tex]\[ 24 \times 1 = 24 \][/tex]
Finally, dividing the products:
[tex]\[ C(10, 6) = \frac{5040}{24} = 210 \][/tex]
Therefore, the value of [tex]\( C(10, 6) \)[/tex] is:
[tex]\[ 210 \][/tex]
This is the number of ways to choose 6 items from a set of 10 items.
