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Hint: From the concept of permutation and combination, if we want to choose ‘r’ things from a total of ‘n’ things (n > r), then the number of ways to do so is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Using this formula, we can solve this question.

Complete step-by-step answer:

Before proceeding with the question, we must know all the formulas that will be required to solve this question.

In permutations and combinations, we have a formula which can be used to find the number of ways in which we can select r things from a total number of n things. This formula is given by,

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}...............\left( 1 \right)$

In the question, it is given that there are 4 officers and 8 privates. We are required to find the number of ways in which we can choose 6 persons such that it includes exactly one officer. Also, we are required to find the number of ways in which we can choose 6 persons such that it includes at least one officer.

(1) In this part, we have to find the number of ways in which we can choose 6 persons such that it includes exactly one officer. So, we can say that among the 6 chosen persons, 1 will be an officer and other five will be privates.

From formula $\left( 1 \right)$, the number of ways in which we can choose 1 officer from a total of 4 officers is equal to,

$\begin{align}

& {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!} \\

& \Rightarrow \dfrac{4\times 3!}{3!} \\

& \Rightarrow 4 \\

\end{align}$

Also, from formula $\left( 1 \right)$, the number of ways in which we can choose 5 privates from a total of 8 privates is equal to,

\[\begin{align}

& {}^{8}{{C}_{5}}=\dfrac{8!}{5!\left( 8-5 \right)!} \\

& \Rightarrow \dfrac{8\times 7\times 6\times 5!}{5!.3!} \\

& \Rightarrow \dfrac{8\times 7\times 6}{3\times 2\times 1} \\

& \Rightarrow 56 \\

\end{align}\]

Since we have to choose 1 officer and 5 privates, the number of ways to do so will be given by multiplying the above two obtained numbers. So, the number of ways in which we can choose 6 persons such that it includes exactly one officer is equal to $56\times 4=224$.

(2) In this part, we have to find the number of ways in which we can choose 6 persons such that it includes at least one officer. The number of ways in which we can choose at least one officer can be found by subtracting the number of ways in which we can choose 6 persons from 4 officers and 8 privates and the number of ways in which we can choose 6 persons such that it includes no officer.

Using formula $\left( 1 \right)$, the number of ways in which we can select 6 persons from 4 officers and 8 privates is equal to,

\[\begin{align}

& {}^{12}{{C}_{6}}=\dfrac{12!}{6!\left( 12-6 \right)!} \\

& \Rightarrow \dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6!.6!} \\

& \Rightarrow \dfrac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1} \\

& \Rightarrow 924 \\

\end{align}\]

The number of ways we can select 6 persons such that it includes no officer will be equal to the number of ways in which we can select 6 privates out of 8 privates. Using formula $\left( 1 \right)$, we get,

\[\begin{align}

& {}^{8}{{C}_{6}}=\dfrac{8!}{6!\left( 8-6 \right)!} \\

& \Rightarrow \dfrac{8\times 7\times 6!}{6!2!} \\

& \Rightarrow 4\times 7 \\

& \Rightarrow 28 \\

\end{align}\]

As discussed in the above paragraph, the number of ways in which we can choose 6 persons such that it includes at least one officer can be found by subtracting the two obtained numbers is equal to $924-28=896$.

Note: There is a possibility that one may commit a mistake while finding the answer of the part (1). There is a possibility that one may add the two obtained numbers instead multiplying them. But since we have to select both 1 officer and 5 privates, we have to multiply the two obtained numbers.

Complete step-by-step answer:

Before proceeding with the question, we must know all the formulas that will be required to solve this question.

In permutations and combinations, we have a formula which can be used to find the number of ways in which we can select r things from a total number of n things. This formula is given by,

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}...............\left( 1 \right)$

In the question, it is given that there are 4 officers and 8 privates. We are required to find the number of ways in which we can choose 6 persons such that it includes exactly one officer. Also, we are required to find the number of ways in which we can choose 6 persons such that it includes at least one officer.

(1) In this part, we have to find the number of ways in which we can choose 6 persons such that it includes exactly one officer. So, we can say that among the 6 chosen persons, 1 will be an officer and other five will be privates.

From formula $\left( 1 \right)$, the number of ways in which we can choose 1 officer from a total of 4 officers is equal to,

$\begin{align}

& {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!} \\

& \Rightarrow \dfrac{4\times 3!}{3!} \\

& \Rightarrow 4 \\

\end{align}$

Also, from formula $\left( 1 \right)$, the number of ways in which we can choose 5 privates from a total of 8 privates is equal to,

\[\begin{align}

& {}^{8}{{C}_{5}}=\dfrac{8!}{5!\left( 8-5 \right)!} \\

& \Rightarrow \dfrac{8\times 7\times 6\times 5!}{5!.3!} \\

& \Rightarrow \dfrac{8\times 7\times 6}{3\times 2\times 1} \\

& \Rightarrow 56 \\

\end{align}\]

Since we have to choose 1 officer and 5 privates, the number of ways to do so will be given by multiplying the above two obtained numbers. So, the number of ways in which we can choose 6 persons such that it includes exactly one officer is equal to $56\times 4=224$.

(2) In this part, we have to find the number of ways in which we can choose 6 persons such that it includes at least one officer. The number of ways in which we can choose at least one officer can be found by subtracting the number of ways in which we can choose 6 persons from 4 officers and 8 privates and the number of ways in which we can choose 6 persons such that it includes no officer.

Using formula $\left( 1 \right)$, the number of ways in which we can select 6 persons from 4 officers and 8 privates is equal to,

\[\begin{align}

& {}^{12}{{C}_{6}}=\dfrac{12!}{6!\left( 12-6 \right)!} \\

& \Rightarrow \dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6!.6!} \\

& \Rightarrow \dfrac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1} \\

& \Rightarrow 924 \\

\end{align}\]

The number of ways we can select 6 persons such that it includes no officer will be equal to the number of ways in which we can select 6 privates out of 8 privates. Using formula $\left( 1 \right)$, we get,

\[\begin{align}

& {}^{8}{{C}_{6}}=\dfrac{8!}{6!\left( 8-6 \right)!} \\

& \Rightarrow \dfrac{8\times 7\times 6!}{6!2!} \\

& \Rightarrow 4\times 7 \\

& \Rightarrow 28 \\

\end{align}\]

As discussed in the above paragraph, the number of ways in which we can choose 6 persons such that it includes at least one officer can be found by subtracting the two obtained numbers is equal to $924-28=896$.

Note: There is a possibility that one may commit a mistake while finding the answer of the part (1). There is a possibility that one may add the two obtained numbers instead multiplying them. But since we have to select both 1 officer and 5 privates, we have to multiply the two obtained numbers.

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