A club with six members is to choose three​ officers: ​ president, vice-president, and​ secretary-treasurer. If each office is to be held by one person and no person can hold more than one​ office, in how many ways can those offices be​ filled?

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Answer:

120 ways

Step-by-step explanation:

We have a total of 6 people, and we want to form groups of 3, so we can do a combination of 6 choose 3.

But inside a group of 3 people, we need to choose the office of each one, and the number of possibilities for this is calculated using factorial.

So, we have the following:

Number of groups of 3 among 6 people:

C(6,3) = 6!/(3!*3!) = 6*5*4/6 = 20

Different offices inside the group of 3:

3! = 3*2 = 6

Then, to find the total number of possibilities that offices can be filled, we multiply these results:

20 * 6 = 120

(This problem can also be solved using permutation, as the order of the elements in the group matters:

P(6,3) = 6!/3! = 6*5*4 = 120)

In 120 ways can those offices be​ filled.

Given that,

A club with six members is to choose three​ officers:

If each office is to be held by one person and no person can hold more than oneoffice.

We have to find,

In how many ways can those offices be​ filled.

According to the question,

Total number of members = 6

Total number group want to form = 3

Then, It is a combination of 6 to choose 3.

Inside a group of 3 people, choose the office of each one, and the number of possibilities for this is calculated using factorial.

Therefore,

Number of groups of 3 among 6 people:

[tex]^6C_3 = \frac{6!}{(6-3!)3!} = \frac{720}{36} = 20[/tex]

Different offices inside the group of 3:

[tex]3! = 3\times 2 \times1 = 6[/tex]

The total number of possibilities that offices can be filled, we multiply these results:

[tex]= 20 \times 6 = 120[/tex]

By using permutation, as the order of the elements in the group matters:

[tex]^6P_3 = \dfrac{6!}{3!} = 6\times5\times4 = 120[/tex]

Hence, In 120 ways can those offices be​ filled.

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