Question

In: Statistics and Probability

Q.1 A security system uses a 2-letter password, but no letter can be used more than...

Q.1 A security system uses a 2-letter password, but no letter can be used more than once. How many possible passwords are there if 15 of the letters of the alphabet can be used by the system? (Use these numbers for this question only!)

For this question indicate only what the value of N is?

Q2.

A security system uses a 5-letter password, but no letter can be used more than once. How many possible passwords are there if 25 of the letters of the alphabet can be used by the system? (Use these numbers for this question only!)

For this question indicate only what the value of r is?

Q3.

A security system uses a four-letter password, but no letter can be used more than once. How many possible passwords are there if 22 of the letters of the alphabet can be used by the system? (Use these numbers for this question only!)

Expert Solution

Q.1 A security system uses a 2-letter password, but no letter can be used more than once. How many possible passwords are there if 15 of the letters of the alphabet can be used by the system?

At first place 15 options and at 2nd place 14 options

Total number of cases = 15*14 = 210

A security system uses a 5-letter password, but no letter can be used more than once. How many possible passwords are there if 25 of the letters of the alphabet can be used by the system?

At first place 25 options, at 2nd place 24 options, at 3rd 23, at 4th 22 and at 5th 21

Total number of cases = 25*24*23*22*21 = 6375600

Alternate

This is a problem of number of arrangement without repetition and order does matter . we will use permutation.

n = 25 r = 5

Total number of cases = nPr = 25P5 = 25!/(25-5)! = 25!/20! = 25*24*23*22*21 = 6375600

Q3.

A security system uses a four-letter password, but no letter can be used more than once. How many possible passwords are there if 22 of the letters of the alphabet can be used by the system?

At first place 24 options, at 2nd place 23 options, at 3rd 22 andat 5th 21

Total number of cases = 24*23*22*21 = 255024

Alternate

This is a problem of number of arrangement without repetition and order does matter . we will use permutation.

n = 24 r = 4

Total number of cases = nPr = 24P4 = 24!/(24-4)! = 24!/20! = 24*23*22*21 = 255024

orchestra answered 1 year ago

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