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Ahat [919]
3 years ago
5

A club with 16 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?
Mathematics
1 answer:
shepuryov [24]3 years ago
6 0

Answer:

The offices can be filled in 3,360 ways

Step-by-step explanation:

In this question, we are faced with a selection problem.

There are three offices and 3 officers to be selected.

For the first position, we are to select 1 out of 16. Now, for the second position, we are left with 15 members from which we are to select one. This is because the first selection has been picked and cannot partake in the subsequent selections.

Likewise for the third position, we are selecting 1 out of 14

The number of ways this can be done is thus; 16C1 × 15C1 × 14C1 = 16 × 15 × 14 = 3,360 ways

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Step-by-step explanation:

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There are 99 males and 121 females participating in a marathon. What participants are females?
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Answer:

121..??

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Please help this is in my test tomorrow♡
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<u><em>67</em></u>

Step-by-step explanation:

<u><em>50-40+87-30</em></u>

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5 0
3 years ago
A deck of cards contains RED cards numbered 1,2,3,4,5, BLUE cards numbered 1,2,3, and GREEN cards numbered 1,2,3,4,5,6. If a sin
Tatiana [17]

Answer: 1/52

Step-by-step explanation:

A deck of cards contains 52 cards

Red cards are numbered 1,2,3

Blue cards are numbered 1,2,3,4,5,6

Green cards are numbered 1,2

Number of Red card = 3

Number of blue card = 6

Number of Green card = 2

Let Pr(R) = Probability of picking a Red card

Let Pr(B) = probability of picking a blue card

Let Pr(G) = probability of picking a green card

Let Pr(RE) = Probability of picking a red even card

Let Pr(RO) = probability of picking a red odd card

Let Pr(BE) =probability of picking a blue even card

Let Pr(BO) = probability of picking a blue odd card

Let Pr(GE) = probability of picking a green even card

Let Pr(GO) = probability of picking a green odd card

Pr(R) = 3/52

Since we have 3 red cards,

Pr(RE) = 1/3

Pr(RO) =2/3

Pr(B) = 6/52

= 3/26

Since we have 6 blue cards ,

Pr(BE) = 3/6

= 1/2

Pr(BO) = 3/6

= 1/2

Pr (G) = 2/52

= 1/26

Since we have 2 green cards,

Pr(GE) = 1/2

Pr(GO) = 1/2

The probability of picking a Green card and an odd green card is

Pr(G) n pr(GO)

1/26 * 1/2

= 1/52

5 0
3 years ago
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