Question

Number of elements in set of values of $r$ for which ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}\ge {}^{20}{{C}_{13}}$ is satisfied.
A. 4
B. 5
C. 7
D. 10

Answer 1

We first find the right substitution for ${}^{n}{{C}_{m}}+{}^{n}{{C}_{m-1}}={}^{n+1}{{C}_{m}}$. We use the identity to find the simplified form of ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}$. Then we use the increasing and decreasing value to find the values for $r$ which satisfies ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}\ge {}^{20}{{C}_{13}}$.

Complete step by step solution:
In this problem, we use the relation of combinations where ${}^{n}{{C}_{m}}+{}^{n}{{C}_{m-1}}={}^{n+1}{{C}_{m}}$.
We complete the summation in the left part of the ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}\ge {}^{20}{{C}_{13}}$.
We take two at a time, breaking the middle one.
So, ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}=\left( {}^{18}{{C}_{r-2}}+{}^{18}{{C}_{r-1}} \right)+\left( {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}} \right)$.
For $\left( {}^{18}{{C}_{r-2}}+{}^{18}{{C}_{r-1}} \right)$, we apply $n=18,m=r-1$. We get ${}^{18}{{C}_{r-2}}+{}^{18}{{C}_{r-1}}={}^{19}{{C}_{r-1}}$.
For $\left( {}^{18}{{C}_{r}}+{}^{18}{{C}_{r-1}} \right)$, we apply $n=18,m=r$. We get ${}^{18}{{C}_{r}}+{}^{18}{{C}_{r-1}}={}^{19}{{C}_{r}}$.
So, ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}={}^{19}{{C}_{r-1}}+{}^{19}{{C}_{r}}$.
For ${}^{19}{{C}_{r-1}}+{}^{19}{{C}_{r}}$, we apply $n=19,m=r$. We get ${}^{19}{{C}_{r-1}}+{}^{19}{{C}_{r}}={}^{20}{{C}_{r}}$.
Therefore, ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}={}^{20}{{C}_{r}}$. We get ${}^{20}{{C}_{r}}\ge {}^{20}{{C}_{13}}$.
Now we know that ${}^{n}{{C}_{m}}={}^{n}{{C}_{n-m}}$. So, to discuss $r$ in ${}^{20}{{C}_{r}}\ge {}^{20}{{C}_{13}}$, we use the values from 10 to 20.
We get ${}^{20}{{C}_{20}}=1,{}^{20}{{C}_{19}}=20,{}^{20}{{C}_{18}}=190,....$.
So, we can see the more we get towards 10, the value gets bigger than the previous one.
Therefore, ${}^{20}{{C}_{13}}={}^{20}{{C}_{13}}$, ${}^{20}{{C}_{12}}\ge {}^{20}{{C}_{13}}$, ${}^{20}{{C}_{11}}\ge {}^{20}{{C}_{13}}$, ${}^{20}{{C}_{10}}\ge {}^{20}{{C}_{13}}$.
All these values of $r=10,11,12,13$ satisfies ${}^{18}{{C}_{r-2}}+2\times {}^{18}{{C}_{r-1}}+{}^{18}{{C}_{r}}\ge {}^{20}{{C}_{13}}$.
The correct option is A.

Note: There are some constraints in the form of ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. Also, we need to remember the fact that the notion of choosing r objects out of n objects is exactly equal to the notion of choosing $\left( n-r \right)$ objects out of n objects. The mathematical expression is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}={}^{n}{{C}_{n-r}}$.