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

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3 years ago