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gogolik [260]
2 years ago
12

How many different ways are there to arrange the letters of the word PRISM?

Mathematics
1 answer:
Advocard [28]2 years ago
6 0

Answer:

5! (5 factorial) or 120 ways

Step-by-step explanation:

You can arrange P different ways. R can be arranged only 4 different ways because Pwill be in one of the spots. Next, you can arrange "I" 3 different ways because P and R are already arranged in 2 spots, meaning you cant put "I" there. Next, S can be arranged in 2 different ways for the same reason. And lastly, M can only be arranged in 1 different way because the other spots are taken.

You can put it in any order. For example, M could be arranged 5 ways and P could be arranged in 1 way. You will still get 120 ways

Hopefully this helped. The explanation might be a bit confusing but I tried my best. I'm not good with explaining my solutions but I always try

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Five computer program modules are ranked as M1, M2, M3, M4, and M5 according to the ascending order of effort required to debug
gulaghasi [49]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Technician selects three out of 5 systems  

In C(5,3)=10ways, this can be achieved  

In part a:

Space sample chooses 3 of a 5 systems  

(M_1, \ M_2,\ M_3),(M_1,M_2,M_4) \ (M_1,M_2,M_5) \ (M_1,M_3,M_4) \ (M_1,M_3,M_5),(M_1,M_4,M_5) \ (M_2,M_3,M_4)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5)}

In point b:

A =MODULE WHICH INCLUDE M1 minimal amount of effort  

Outcomes probable =

(M_1,M_2,M_3),\ (M_1,M_2,M_4) \ (M_1,M_2,M_5)\ (M_1,M_3,M_4)\\\\(M_1,M_3,M_5),\ (M_1,M_4,M)5)\ =\ 6

\to p(A)=\frac{6}{10}\\\\

            =0.6

In point c:

B = highest effort that is M_5

Potential result=

(M_1,M_2,M_5) \ (M_1,M_3,M_5) \ (M_2,M_3,M_5)\(M_2,M_4,M_5) \\ (M_2,M_4,M_5), \ (M_3,M_4,M_5) \ =\ 6  \\\\

\to B= \frac{6}{10} \\\\

        =0.6

\to P(B)=10

In point d:

\to \ A  \ intersection \ B=(M_1,M_2,M_5), \ (M_1,M_3,M_5) \ ,(M_1,M_4,M_5)

\to A (A \ intersection \ B) = \frac{3}{10} \\\\\ \ \ \ \ \

                                      =0.3

In point e:

\to (A \cup B) =  (M_1,M_2,M_3),\ (M_1,M_2,M_4)\ (M_1,M_2,M_5)(M_1,M_3,M_4)\ (M_1,M_3,M_5), \\ (M_1,M_4,M_5)\ (M_2,M_3,M_5) \ (M_2,M_4,M_5),(M_3,M_4,M_5) \ = \ 9\to P(A \cap B)=\frac{9}{10}

                    = 0.9

In point f:

\to (A\cap B) = \frac{3}{10}

                 = 0.3

In point g:

\to (A \cup B) = \frac{7}{10}

                 =0.7

In point h:

\to p(A \cap B) = 0.3 \neq 0

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y-2= -2x-12

y= -2x-10
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Answer:

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