19 June, 13:02

# 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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1. 19 June, 14:15
0
120 ways

Step-by-step explanation:

Total numbers of members 'n' = 6

number of officers to choose 'r' = 3

So, for the number of permutation of (n=6) members taken (r=3) at a time, the formula is

nPr = n! / (n-r) !

6P3 = 6! / (6-3) !

6P3 = 6! / 3! = > (6x 5x4x3x2x1) / (3x2x1)

6P3 = 120

Thus, those officers can be filled in 120 ways
2. 19 June, 14:25
0
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)