Question

The supreme court has given a $6$ to $3$ decision upholding a lower court. The number of ways it can give a majority decision reversing the lower court is
A. $256$
B. $276$
C. $245$
D. $226$

Answer 1

In this problem we need to find the number of ways to lower the decision. In the given problem we have given that the supreme court has given a $6$ to $3$ decision upholding a lower court. From this we will calculate the number of judges in the court. To lower the decision, we need to have more than half of the judge's support. So, we will calculate the number of required judges to lower the decision. Now we will calculate the number of combinations that satisfies the requirement which is our required result.

Complete step by step answer:
Given that, the supreme court has given a $6$ to $3$ decision upholding a lower court.
Total number of judges in the court is given by $6+3=9$.
In the court we have $9$. To lower the decision, we need to have more than $50\%$ of the judge’s support. $50\%$ of $9$ means $4.5$ that means we need to have minimum $5$ judge’s support.
Now the number of ways that minimum $5$ judges lower the decision out of $9$ judges is given by
$N={}^{9}{{C}_{9}}+{}^{9}{{C}_{8}}+{}^{9}{{C}_{7}}+{}^{9}{{C}_{6}}+{}^{9}{{C}_{5}}$
Simplifying the above equation by using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, then we will get
$\begin{aligned} & N=\dfrac{9!}{(9-9)!9!}+\dfrac{9!}{(9-8)!8!}+\dfrac{9!}{(9-7)!7!}+\dfrac{9!}{(9-6)!6!}+\dfrac{9!}{(9-5)!5!} \\\ & \Rightarrow N=\dfrac{9!}{0!9!}+\dfrac{9!}{1!8!}+\dfrac{9!}{2!7!}+\dfrac{9!}{3!6!}+\dfrac{9!}{4!5!} \\\ \end{aligned}$
Simplifying the above equation by substituting all the known values, then we will get
$N=\dfrac{362880}{1\times 362880}+\dfrac{362880}{1\times 40320}+\dfrac{362880}{2\times 5040}+\dfrac{362880}{6\times 720}+\dfrac{362880}{24\times 120}$
Simplifying the above equation by using basic mathematical operations, then we will have
$\begin{aligned} & N=1+9+36+84+126 \\\ & \Rightarrow N=256 \\\ \end{aligned}$

So, the correct answer is “Option A”.

Note: We have calculated the factorial values of any number by using the formula $n!=n\left( n-1 \right)\left( n-2 \right)......3\times 2\times 1$. We can also use some basic permutation and combination formulas like ${}^{n}{{C}_{n}}=1$, ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$.