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kotykmax [81]
2 years ago
8

A lock has a three number code made up of 17 numbers if none of the numbers are allowed to repeat how many different ways can yo

u choose three different numbers in order for unit code?
Mathematics
1 answer:
Marrrta [24]2 years ago
8 0

Answer:

680

Step-by-step explanation:

Use the binomial coefficient where you choose k=3 numbers out of n=17 possible numbers and find the total amount of combinations since order does not matter:

\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}\\ \\\binom{17}{3}=\frac{17!}{3!(17-3)!}\\\\\binom{17}{3}=\frac{17!}{3!(14)!}\\\\\binom{17}{3}=\frac{17*16*15}{3*2*1}\\\\\binom{17}{3}=680

Thus, you can make 680 three-non-repeating-number codes

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Nataly [62]

Answer:

the two numbers are 16 and -25

Step-by-step explanation:

Let the two numbers be x and y

x+y=-9

x-y=-41

By substitution

y=-9-x

x-(-9-x)=-41

x+9+x=-41

2x=-41-9

2x=-50

x=-25

y=-9-(-25)

y=16

6 0
3 years ago
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Suppose :

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_____________________________

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=====》

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&

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Thus :

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3^y = 2 ===》 y = Log _ 3 { 2 }

8 0
3 years ago
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nikitadnepr [17]

Answer:

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6 0
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Zigmanuir [339]

Answer:

5\frac{1}{9}

Step-by-step explanation:

When dividing 46 by 9 quotient is 5 and remainder is 46 - 45 = 1

5\frac{1}{9}

5 0
3 years ago
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​

Step-by-step explanation:

6 0
3 years ago
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