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?

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 one office.

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:

Different offices inside the group of 3:

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

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

Hence, In 120 ways can those offices be filled.

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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)

72 fudge and 75 toffee

step-by-step explanation:

i hope this : )

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2