Question

Based on this information answer the questions given below.

(i) ${}^{n}{{P}_{r}}=r! {}^{n}{{C}_{r}}$

(ii) ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$

What is the value of ${}^{8}{{C}_{4}}+{}^{8}{{C}_{3}}$?

(a) ${}^{8}{{C}_{3}}$

(b) 63

(c) 35

(d) ${}^{9}{{C}_{4}}$

Answer 1

We will first find ${}^{8}{{C}_{4}}$ and ${}^{8}{{C}_{3}}$ using the combination formula ${}^{n}{{C}_{r}}=\dfrac{{}^{n}{{P}_{r}}}{r!}=\dfrac{n!}{r!(n-r)!}$ and then add them together to get the value and then we will find the same value using the formula ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ mentioned in the question and will finally verify both.

Complete step-by-step answer:

Before proceeding with the question, we should know about permutations and combinations.

Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group it is said to be permutations whereas the order in which they are represented is called combination.

A permutation is the choice of r things from a set of n things without replacement and where the order matters. ${}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}..........(1)$

A combination is the choice of r things from a set of n things without replacement and where order doesn't matter. ${}^{n}{{C}_{r}}=\dfrac{{}^{n}{{P}_{r}}}{r!}=\dfrac{n!}{r!(n-r)!}......(2)$

Now first solving ${}^{8}{{C}_{4}}$ by using combination formula from equation (2) we get,

${}^{8}{{C}_{4}}=\dfrac{8!}{4!\times (8-4)!}=\dfrac{8\times 7\times 6\times 5\times 4!}{4!\times 4!}=\dfrac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=70......(3)$

Now first solving ${}^{8}{{C}_{3}}$ by using combination formula from equation (2) we get,

${}^{8}{{C}_{3}}=\dfrac{8!}{3!\times (8-3)!}=\dfrac{8\times 7\times 6\times 5!}{3!\times 5!}=\dfrac{8\times 7\times 6}{3\times 2\times 1}=56......(4)$

Now adding equation (3) and equation (4) to get ${}^{8}{{C}_{4}}+{}^{8}{{C}_{3}}$,

${}^{8}{{C}_{4}}+{}^{8}{{C}_{3}}=70+56=126.....(5)$

Given from the following passage in the question ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}........(6)$. Considering n as 8 and r as 4 and substituting this in equation (5) we get,

${}^{8}{{C}_{4}}+{}^{8}{{C}_{3}}={}^{9}{{C}_{4}}........(7)$

Now solving ${}^{9}{{C}_{4}}$in equation (7) to verify if it is equal to 126 or not.

${}^{9}{{C}_{4}}=\dfrac{9!}{4!\times (9-4)!}=\dfrac{9\times 8\times 7\times 6\times 5!}{4!\times 5!}=\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}=126......(8)$

So we see from equation (7) and equation (8) that ${}^{9}{{C}_{4}}$ is the answer.

So hence option (d) is the right answer.

Note: Remembering the formula of combination mentioned in equation (2) in the solution is the key here. We can make a mistake in expanding the factorial in a hurry so we need to be careful with this step.