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arlik [135]
3 years ago
10

Four members from a ​"55"person committee are to be selected randomly to serve as​ chairperson, vice-chairperson,​ secretary, an

d treasurer. The first person selected is the​ chairperson; the​ second, the​ vice-chairperson; the​ third, the​ secretary; and the​ fourth, the treasurer. How many different leadership structures are​ possible?
Mathematics
1 answer:
liq [111]3 years ago
3 0

Answer:

8,185,320 different leadership structures

Step-by-step explanation:

Since the order at which the members of the committee are chosen matters, this is a permutation of 4 out 55 people and it is given by:

n=\frac{55!}{(55-4)!}=55*54*53*52 \\\\n=8,185,320

8,185,320 different leadership structures are​ possible.

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common difference: d = -14 (we decrease by 14 each time, eg, 151-14 = 137)

nth term of this arithmetic sequence is...
an = a1+d(n-1)
an = 151+(-14)(n-1)
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This will be used in the formula below

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</span>S26 = -624

The final answer here is choice C) -624
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