To solve the given expression for [tex]\( n = 5 \)[/tex] and [tex]\( r = 2 \)[/tex], we will break down the problem step by step to find the value of [tex]\( x \)[/tex].
The expression to evaluate is:
[tex]\[
r! \times \left( \binom{n-1}{r-1} + \binom{n-1}{r} \right) + (r+1)! \times \left( \binom{n-1}{r} + \binom{n-1}{r+1} \right) = x
\][/tex]
First, we need to calculate the factorials and binomial coefficients involved.
1. Factorials:
- [tex]\( r! \)[/tex]:
[tex]\[
r! = 2! = 2 \times 1 = 2
\][/tex]
- [tex]\( (r+1)! \)[/tex]:
[tex]\[
(r+1)! = 3! = 3 \times 2 \times 1 = 6
\][/tex]
2. Binomial Coefficients:
- [tex]\( \binom{n-1}{r-1} \)[/tex]:
[tex]\[
\binom{n-1}{r-1} = \binom{4}{1} = \frac{4!}{1!(4-1)!} = \frac{4!}{1! \cdot 3!} = \frac{4 \times 3!}{1 \times 3!} = 4
\][/tex]
- [tex]\( \binom{n-1}{r} \)[/tex]:
[tex]\[
\binom{n-1}{r} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6
\][/tex]
- [tex]\( \binom{n-1}{r+1} \)[/tex]:
[tex]\[
\binom{n-1}{r+1} = \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3! \cdot 1!} = \frac{24}{6} = 4
\][/tex]
3. First Term:
- Calculate [tex]\( \binom{n-1}{r-1} + \binom{n-1}{r} \)[/tex]:
[tex]\[
\binom{4}{1} + \binom{4}{2} = 4 + 6 = 10
\][/tex]
- Multiply by [tex]\( r! \)[/tex]:
[tex]\[
r! \times \left( \binom{n-1}{r-1} + \binom{n-1}{r} \right) = 2 \times 10 = 20
\][/tex]
4. Second Term:
- Calculate [tex]\( \binom{n-1}{r} + \binom{n-1}{r+1} \)[/tex]:
[tex]\[
\binom{4}{2} + \binom{4}{3} = 6 + 4 = 10
\][/tex]
- Multiply by [tex]\( (r+1)! \)[/tex]:
[tex]\[
(r+1)! \times \left( \binom{n-1}{r} + \binom{n-1}{r+1} \right) = 6 \times 10 = 60
\][/tex]
5. Total Value [tex]\( x \)[/tex]:
- Add the first term and the second term:
[tex]\[
x = 20 + 60 = 80
\][/tex]
Therefore, the value of [tex]\( x \)[/tex] is [tex]\( 80 \)[/tex].
