Question

A corporation must appoint a​ president, chief executive officer​ (CEO), chief operating officer​ (COO), and chief financial officer​ (CFO). It must also appoint a planning committee with threethree different members. There are 1313 qualified​ candidates, and officers can also serve on the committee. Complete parts​ (a) through​ (c) below.

(a) How many different ways can the officers be​ appointed?
(b) How many different ways can the committee be​ appointed?
(c) What is the probability of randomly selecting the committee members and getting the four youngest of the qualified​ candidates?

Answer 1

Answer:

Step-by-step explanation:

There are one president, CEO, COO and CFO.

So 4 executives have to be selected out of the 13 qualified candidates.

a) Ways officers can be appointed = $13C3 = \frac{13(12)(11)}{3(2)(1)} \\=286$

b) Committee consists of 3 different members.

Ways of forming committee =$13C3=286$

c)  Prob of randomly selecting the committee members and getting the four youngest of the qualified​ candidates

Answer 2

The corporation appointing presidents, officers and members of committee is an illustration of combination or selection.

• There are 1716 ways of appointing officers
• There are 286 ways of appointing committee members
• The probability of selecting 4 youngest candidates from three different members is 0.

Given that:

$n = 13$ ---- qualified candidates and officers

(a) Ways of appointing the officers.

From the given details, we understand that there are 3 different officers to be appointed; the CEO, the COO and CFO.

The first can be appointed from the 13 qualified candidates

The second, from the remaining 12

The third, from the remaining 11

So, the number of ways is:

$Ways= 13\times 12\times 11$

$Ways= 1716$

There are 1716 ways of appointing officers

(b) Ways of appointing the committee members.

There are 13 qualified candidates, of which any 3 can be a member of the committee

The number of ways is calculated using the following combination formula:

$Ways = ^nC_r$

Where

$n = 13 \\ r = 3$

So, we have:

$Ways = ^{13}C_3$

$Ways = \frac{13!}{(13 - 3)! \times 3!}$

$Ways = \frac{13!}{10! \times 3!}$

Expand

$Ways = \frac{13\times 12 \times 11 \times 10!}{10! \times 3!}$

$Ways = \frac{13\times 12 \times 11}{3 \times 2 \times 1}$

$Ways = \frac{1716}{6}$

$Ways = 286$

There are 286 ways of appointing committee members

(c) Probability of selecting 4 youngest candidates

From the question, we understand that the member of the committee are 3. This means that it is impossible to get 4 youngest candidates from 3 committee members.

Hence, the probability is 0.

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