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Elena-2011 [213]
3 years ago
13

Jim wants to make a 4 letter password. He cannot repeat any letters in his password.How many combinations can he make

Mathematics
2 answers:
9966 [12]3 years ago
8 0

My previous answer was incorrect, sorry for that... the correct answer would be 358,800

Andreas93 [3]3 years ago
3 0

\huge\text{Hey there!}

\large\text{He has 26 letters all together (A  - Z)}

\large\text{25 if he  starts  from his  second letter (B - Z)}

\large\text{24 if he starts from his third letter (C - Z)}

\large\text{ 23 if he starts from his fourth letter (D - Z)}

\mathsf{26\times25\times24\times23}

\mathsf{26\times 25 = \bf 650}

\mathsf{650\times24\times23}

\mathsf{650\times24 = \bf 15,600}

\mathsf{15,600\times 23}

\mathsf{  =\bf  358,800}

\boxed{\boxed{\large\text{Answer: \huge \bf 358,800}}}\huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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Hello! I want to say that I am sorry if I said/did anything to be rude to you guys. So, I will answer this question in a way of saying sorry.

The correct answer is Option C. 17.

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Mice21 [21]

I'm not 100% positive, but from my calculatons, the answer is C.

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PolarNik [594]

Answer:

4

Step-by-step explanation:

because you do 2-1 and then 3 * 3 the equation would look like this

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3 years ago
In one​ town, 66% of adults have health insurance. What is the probability that 4 adults selected at random from the town all ha
LiRa [457]

Answer:

Probability that randomly selected 4  persons all have insurance is = 0.1897

Step-by-step explanation:

Given that the probability of having an insurance equals 66%=0.66

Thus

Probability that person A has insurance = 0.66

Probability that person B has insurance = 0.66

Probability that person C has insurance = 0.66

Probability that person D has insurance = 0.66

Probability that randomly selected 4  persons all have insurance is

P(A)\times P(B)\times P(C)\times P(D)\\\\=0.66\times 0.66\times 0.66\times 0.66=0.1897

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Arte-miy333 [17]

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