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tigry1 [53]
3 years ago
6

The digits 1-5 are used for a set of locker codes. Suppose the digits cannot repeat. Find the number of possible two digit and t

hree digit codes. Describe any pattern and use it to predict the number of possible five digit codes.
Mathematics
2 answers:
Archy [21]3 years ago
7 0
So you would do
5×4×3= 60
and
5×4=20
so, for a 3 digit code there are 60 possibilities and for a 2 digit code there are 20 possibilities.
TiliK225 [7]3 years ago
5 0

Answer: Number of possible two digit codes= 20

Number of possible three digits code = 60

To make five digit codes the pattern will be given by using permutations:-

^5P_5=\frac{5!}{(5-5)!}

Number of possible five digit codes =120

Step-by-step explanation:

Given: The number of digits used for a set of locker codes =5

If repetition is not allowed, then the number of possible two digit codes is given by by using permutations:-

^5P_2=\frac{5!}{(5-2)!}=\frac{5\times4\times3!}{3!}=20

Similarly, the number of possible three digit codes is given by :-

^5P_3=\frac{5!}{(5-3)!}=\frac{5\times4\times3\times2!}{2!}=60

Similarly, to make five digit codes the pattern will be given by using permutations:-

^5P_5=\frac{5!}{(5-5)!}

Therefore, the number of possible five digit codes =\frac{5!}{(5-5)!}=5!=120

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fenix001 [56]

Answer:

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Step-by-step explanation:

The difference between consecutive integers is 1.

The difference between consecutive odd integers is 2.

Let the smallest odd integer be x.

Then the next greater one is x + 2. The greatest one is x + 4.

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4 0
3 years ago
1. A = 1/2bh
OleMash [197]
1 and 2 i can't answer without proper data.
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