To find the number of possible combinations when selecting [tex]\( r \)[/tex] objects from [tex]\( n \)[/tex] items, we use the combination formula:
[tex]\[ {}_nC_r = \frac{n!}{r!(n-r)!} \][/tex]
Here, [tex]\( n! \)[/tex] represents the factorial of [tex]\( n \)[/tex], which is the product of all positive integers from 1 to [tex]\( n \)[/tex]. Similarly, [tex]\( r! \)[/tex] is the factorial of [tex]\( r \)[/tex], and [tex]\( (n-r)! \)[/tex] is the factorial of the difference between [tex]\( n \)[/tex] and [tex]\( r \)[/tex].
Let's break down the calculation with the provided values:
1. Compute [tex]\( n! \)[/tex]:
For [tex]\( n = 10 \)[/tex],
[tex]\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \][/tex]
2. Compute [tex]\( r! \)[/tex]:
For [tex]\( r = 3 \)[/tex],
[tex]\[ 3! = 3 \times 2 \times 1 = 6 \][/tex]
3. Compute [tex]\( (n - r)! \)[/tex]:
For [tex]\( n - r = 10 - 3 = 7 \)[/tex],
[tex]\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5,040 \][/tex]
4. Compute the number of combinations:
Using the combination formula:
[tex]\[ {}_{10}C_3 = \frac{10!}{3!(10-3)!} = \frac{3,628,800}{6 \times 5,040} \][/tex]
5. Simplify the expression:
[tex]\[
\frac{3,628,800}{30,240} = 120
\][/tex]
Therefore, the number of possible combinations when taking 3 objects from 10 items is [tex]\( 120 \)[/tex].
So, the correct formula and value are:
[tex]\[
{}_n C _r = \frac{n!}{(n-r)!r!} = 120
\][/tex]
