Answer :
To evaluate \({ }_8 C_2\) (often read as "8 choose 2"), we use the combination formula, which is:
[tex]\[ { }_n C_r = \frac{n!}{r! (n-r)!} \][/tex]
In this case, \(n = 8\) and \(r = 2\). Plugging these values into the formula, we get:
[tex]\[ { }_8 C_2 = \frac{8!}{2! (8-2)!} \][/tex]
First, we calculate the factorials involved:
- \(8!\) (8 factorial) is \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320\)
- \(2!\) (2 factorial) is \(2 \times 1 = 2\)
- \((8-2)!\), which is \(6!\) (6 factorial), is \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)
Then, substitute these values into the combination formula:
[tex]\[ { }_8 C_2 = \frac{40320}{2 \times 720} \][/tex]
Next, we simplify the expression by performing the multiplication in the denominator:
[tex]\[ 2 \times 720 = 1440 \][/tex]
So, our fraction becomes:
[tex]\[ { }_8 C_2 = \frac{40320}{1440} \][/tex]
Now, we perform the division:
[tex]\[ \frac{40320}{1440} = 28 \][/tex]
Therefore, the simplified answer for \({ }_8 C_2\) is:
[tex]\[ { }_8 C_2 = 28 \][/tex]
[tex]\[ { }_n C_r = \frac{n!}{r! (n-r)!} \][/tex]
In this case, \(n = 8\) and \(r = 2\). Plugging these values into the formula, we get:
[tex]\[ { }_8 C_2 = \frac{8!}{2! (8-2)!} \][/tex]
First, we calculate the factorials involved:
- \(8!\) (8 factorial) is \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320\)
- \(2!\) (2 factorial) is \(2 \times 1 = 2\)
- \((8-2)!\), which is \(6!\) (6 factorial), is \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)
Then, substitute these values into the combination formula:
[tex]\[ { }_8 C_2 = \frac{40320}{2 \times 720} \][/tex]
Next, we simplify the expression by performing the multiplication in the denominator:
[tex]\[ 2 \times 720 = 1440 \][/tex]
So, our fraction becomes:
[tex]\[ { }_8 C_2 = \frac{40320}{1440} \][/tex]
Now, we perform the division:
[tex]\[ \frac{40320}{1440} = 28 \][/tex]
Therefore, the simplified answer for \({ }_8 C_2\) is:
[tex]\[ { }_8 C_2 = 28 \][/tex]
