Answer :
To evaluate the expression [tex]\({ }_7 C _3\)[/tex], we use the combination formula, which is:
[tex]\[ { }_n C _{ r } = \frac{n!}{r!(n-r)!} \][/tex]
Given [tex]\( n = 7 \)[/tex] and [tex]\( r = 3 \)[/tex], we can substitute these values into the formula:
[tex]\[ { }_7 C _3 = \frac{7!}{3!(7-3)!} \][/tex]
First, we calculate the factorials:
- [tex]\( 7! \)[/tex] (7 factorial) is the product of all positive integers up to 7: [tex]\( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)[/tex]
- [tex]\( 3! \)[/tex] (3 factorial) is the product of all positive integers up to 3: [tex]\( 3! = 3 \times 2 \times 1 = 6 \)[/tex]
- [tex]\( (7-3)! = 4! \)[/tex] (4 factorial) is the product of all positive integers up to 4: [tex]\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)[/tex]
Now, substitute these factorial values back into the combination formula:
[tex]\[ { }_7 C _{3} = \frac{5040}{6 \times 24} \][/tex]
Next, calculate the denominator:
[tex]\[ 6 \times 24 = 144 \][/tex]
So the combination becomes:
[tex]\[ { }_7 C _{3} = \frac{5040}{144} \][/tex]
Finally, divide 5040 by 144:
[tex]\[ { }_7 C _{3} = 35 \][/tex]
Thus, the value of [tex]\({ }_7 C _3\)[/tex] is:
[tex]\[ \boxed{35} \][/tex]
[tex]\[ { }_n C _{ r } = \frac{n!}{r!(n-r)!} \][/tex]
Given [tex]\( n = 7 \)[/tex] and [tex]\( r = 3 \)[/tex], we can substitute these values into the formula:
[tex]\[ { }_7 C _3 = \frac{7!}{3!(7-3)!} \][/tex]
First, we calculate the factorials:
- [tex]\( 7! \)[/tex] (7 factorial) is the product of all positive integers up to 7: [tex]\( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)[/tex]
- [tex]\( 3! \)[/tex] (3 factorial) is the product of all positive integers up to 3: [tex]\( 3! = 3 \times 2 \times 1 = 6 \)[/tex]
- [tex]\( (7-3)! = 4! \)[/tex] (4 factorial) is the product of all positive integers up to 4: [tex]\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)[/tex]
Now, substitute these factorial values back into the combination formula:
[tex]\[ { }_7 C _{3} = \frac{5040}{6 \times 24} \][/tex]
Next, calculate the denominator:
[tex]\[ 6 \times 24 = 144 \][/tex]
So the combination becomes:
[tex]\[ { }_7 C _{3} = \frac{5040}{144} \][/tex]
Finally, divide 5040 by 144:
[tex]\[ { }_7 C _{3} = 35 \][/tex]
Thus, the value of [tex]\({ }_7 C _3\)[/tex] is:
[tex]\[ \boxed{35} \][/tex]
