Sure, let's solve the problem step by step.
We are given the equation involving combinations (also known as binomial coefficients):
[tex]\[ {}^n C_1 + {}^n C_{n-1} = 12 \][/tex]
First, we need to recall the definition of combinations. The combination formula is given by:
[tex]\[ {}^n C_r = \frac{n!}{r!(n-r)!} \][/tex]
For [tex]\({}^n C_1\)[/tex], we substitute [tex]\(r = 1\)[/tex]:
[tex]\[ {}^n C_1 = \frac{n!}{1!(n-1)!} = \frac{n}{1} = n \][/tex]
For [tex]\({}^n C_{n-1}\)[/tex], we substitute [tex]\(r = n-1\)[/tex]:
[tex]\[ {}^n C_{n-1} = \frac{n!}{(n-1)!1!} = \frac{n}{1} = n \][/tex]
Now, substitute these expressions back into the given equation:
[tex]\[ {}^n C_1 + {}^n C_{n-1} = n + n = 2n \][/tex]
We know that:
[tex]\[ 2n = 12 \][/tex]
To solve for [tex]\(n\)[/tex], divide both sides of the equation by 2:
[tex]\[ n = \frac{12}{2} \][/tex]
[tex]\[ n = 6 \][/tex]
Therefore, the value of [tex]\(n\)[/tex] is:
[tex]\[ n = 6 \][/tex]
