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In how many ways can a committee, consisting of a chairman, secretary, treasurer and four ordinary members be chosen from eight persons? (Committees with different chairmen, secretaries, treasurers count as different committees)

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Last updated date: 23rd Jul 2024
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Answer
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Hint: We make use of the concept of ‘combinations’ in mathematics to solve this problem. When elements are chosen from the given set of elements without considering the order in which the elements are picked out is said to be a combination. Formula for combination is: ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.

Complete step-by-step answer:
Note down all the given details and let us understand what each detail conveys.
From the question we see that every committee must contain one chairman, a treasurer, a secretary and along with them the rest of the members who are four in number.
That makes a total of seven people in each committee.
But we have a total of eight people to choose such a committee from.
We use ‘combination’ to find out the total number of ways of choosing a committee that satisfies the requirements. This is because, the order in which we choose the people do not matter since once a person is chosen, they cannot be considered as anyone else in the next choice.
Requirements are as follow:
It is a must to have one chairman, one treasurer and one secretary, this means that:
One chairman is chosen from eight people $ \Rightarrow {}^8{C_1}$
Then out of the remaining seven, we choose one treasurer $ \Rightarrow {}^7{C_1}$
Next from the remaining six people, let us take one secretary $ \Rightarrow {}^6{C_1}$
The next mandatory requirement is that four more people need to be there within the committee.
The four members out of the remaining five people can be chosen in this way $ \Rightarrow {}^5{C_4}$
Now that the requirements have been met, we combine all of the members to find the possible number of combinations.
For this we need to multiply all the combinations since the committee definitely has:
One chairman, one treasurer, one secretary and four members
$ \Rightarrow {}^8{C_1} \times {}^7{C_1} \times {}^6{C_1} \times {}^5{C_4}$
Now we know that there can only be seven people chosen from the eight given people so that combination will look like this
$ \Rightarrow {}^8{C_7}$
So the total number of ways in which the selection of committee members can be done is by the selection of required members out of the total selected members.
It can be represented as follows:
$ \Rightarrow \dfrac{{{}^8{C_1} \times {}^7{C_1} \times {}^6{C_1} \times {}^5{C_4}}}{{{}^8{C_7}}}$
 $ \Rightarrow \dfrac{{{}^8{C_1} \times {}^7{C_1} \times {}^6{C_1} \times {}^5{C_4}}}{{{}^8{C_7}}} = \dfrac{{8 \times 7 \times 6 \times 5}}{8}$
$ \Rightarrow \dfrac{{8 \times 7 \times 6 \times 5}}{8} = 210$
But we also need to keep in mind that there are eight different people who can be chosen as the chairman, treasurer or secretary so:
$ \Rightarrow 8 \times 210 = 1680$
Therefore the total number of ways for committee selection is $1680$.
So, the correct answer is “$1680$”.

Note: Keep in mind that the two concepts that we might get confused about are ‘permutation’ and ‘combination’. The process of consideration of order in combinations and permutations is what distinguishes them. We must consider the order in which the elements are arranged in permutations, but we do not need to keep in mind the order of arrangement in combinations.