Question

*PERMUTATIONS*

How many tablets of 3 letters and 3 numbers can we form with the letters {A, B, C, D, E} and the 10 digits if the tablet should have:

◦ Letters and numbers should be kept together without repetition?

◦ Letters and numbers should be kept together with repetition?

◦ Letters and numbers do not have to be kept together without repetition?

◦ Letters and numbers do not have to be kept together with repetition?

Answer 1

(a)

Letters and numbers should be kept together without repetition:

Number of selections of First letter = 5

Number of selections of Second letter = 4

Number of selections of Third letter = 3

Number of selections of First number = 10

Number of selections of Second number = 9

Number of selections of Third number = 8

Number of arrangements of Group of 3 letters and 3 numbers = 2

So,

Total of ways = 5 X 4 X 3 X 10 X 9 X 8 X 2 = 86,400

So,

Answer is:

86,400

(b)

Letters and numbers should be kept together with repetition:

Number of selections of First letter = 5

Number of selections of Second letter = 5

Number of selections of Third letter = 5

Number of selections of First number = 10

Number of selections of Second number = 10

Number of selections of Third number = 10

Number of arrangements of Group of 3 letters and 3 numbers = 2

So,

Total of ways = 5³ X 10³ X 2 = 250,000

So,

Answer is:

250,000

(c)

Letters and numbers do not have to be kept together without repetition:

Number of selections of First letter = 5

Number of selections of Second letter = 4

Number of selections of Third letter = 3

Number of selections of First number = 10

Number of selections of Second number = 9

Number of selections of Third number = 8

Number of arrangements of of 6 characters = 6!

So,

Total of ways = 5 X 4 X 3 X 10 X 9 X 8 X 6! = 31,104,000

So,

Answer is:

31,104,000

(d)

Letters and numbers need not have to be kept together with repetition:

Number of selections of First letter = 5

Number of selections of Second letter = 5

Number of selections of Third letter = 5

Number of selections of First number = 10

Number of selections of Second number = 10

Number of selections of Third number = 10

Number of arrangements of 6 characters = 6!

So,

Total of ways = 5³ X 10³ X 6! = 90,000,000

So,

Answer is:

90,000,000