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(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

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Hint: 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,

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

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,

\[{}^{8}{{C}_{4}}+{}^{8}{{C}_{4}}={}^{9}{{C}_{4}}........(7)\]

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.

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,

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

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,

\[{}^{8}{{C}_{4}}+{}^{8}{{C}_{4}}={}^{9}{{C}_{4}}........(7)\]

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.