To determine the number of combinations without repetition for [tex]\( n = 4 \)[/tex] and [tex]\( r = 3 \)[/tex], we use the combinations formula, which is also known as "n choose r" or denoted as [tex]\( \binom{n}{r} \)[/tex].
The formula for combinations is given by:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
Here's a detailed, step-by-step solution:
1. Plug in the values of [tex]\( n \)[/tex] and [tex]\( r \)[/tex]:
[tex]\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} \][/tex]
2. Simplify inside the factorial expressions:
[tex]\[ (4 - 3) = 1 \][/tex]
[tex]\[ \binom{4}{3} = \frac{4!}{3! \cdot 1!} \][/tex]
3. Calculate the factorials:
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]
[tex]\[ 3! = 3 \times 2 \times 1 = 6 \][/tex]
[tex]\[ 1! = 1 \][/tex]
4. Substitute the factorial values into the formula:
[tex]\[ \binom{4}{3} = \frac{24}{6 \times 1} \][/tex]
5. Simplify the division:
[tex]\[ \binom{4}{3} = \frac{24}{6} = 4 \][/tex]
Therefore, the number of combinations without repetition when [tex]\( n = 4 \)[/tex] and [tex]\( r = 3 \)[/tex] is [tex]\( 4 \)[/tex].
The correct answer is:
4
