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Damm [24]
3 years ago
14

If c is a set with c elements how many elements are in the power set of c

Mathematics
1 answer:
Fofino [41]3 years ago
3 0

There would be 2 raised to c elements in the power set. Elements of the power set are formed by making a binary decision on each element of the input set: include or do not include. Since there are two choices for each element of the input set, and there are c elements in the input set, there are 2^c possible arrangements of these input elements.

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Read 2 more answers
A friend of mine is giving a dinner party. His current wine supply includes 9 bottles of zinfandel, 10 of merlot, and 12 of cabe
e-lub [12.9K]

Answer:

Step-by-step explanation:

From the given information:

The total number of wine = 9 + 10 + 12 = 31

(1)

The number of distinct sequences used for serving any five wines can be estimated by using the permutation of the number of total wines with the number of wines served.

i.e

= ^{31}P_5

=\dfrac{31!}{(31-5)!}

=\dfrac{31!}{(26)!}

=\dfrac{31\times 30\times 29\times 28\times 27\times 26!}{(26)!}

= 20389320

(2)

If the first two wines served = zinfandel and the last three is either merlot or cabernet;

Then, the no of ways we can achieve this is:

= ^9P_2\times ^{22}P_3

= \dfrac{9!}{(9-2)!}\times \dfrac{22!}{(22-3)!}

= \dfrac{9!}{(7)!}\times \dfrac{22!}{(19)!}

= \dfrac{9*8*7!}{(7)!}\times \dfrac{22*21*20*19!}{(19)!}

= 665280

(3)

The probability that no zinfandel is served is computed as follows:

Total wines (with zinfandel exclusion) = 31 - 9 = 22

Now;

the required probability is:

= \dfrac{^{22}P_5 }{^{31}P_5}

= \dfrac{\dfrac{22!}{(22-5)!} } {\dfrac{31!}{(31-5)!} }

= \dfrac{\dfrac{22!}{17!} } {\dfrac{31!}{(26)! }}

= \dfrac{(\dfrac{22*21*20*19*18*17!}{17!})} {(\dfrac{31*30*29*28*27*26!}{(26)! })}

= 0.1549

≅ 0.155

4 0
3 years ago
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