Mathematics Probability

Question

A password is a sequence of letters (a{z) and digits (0{9). Find the number of passwords of length 10 under the constraints in (a), (b) or (c) (three separate problems). Express your answer using factorials and integers, products and ratios of them, and/or sums of such things.

(a) There are 3 letters and 7 digits, and at most one `9'.

(b) There are 6 letters and 4 digits, and no digit occurs twice.

(c) No letters are used BUT the rst four digits are not all the same AND the last

four digits are not all the same, AND the 4th, 5th, 6th and 7 digits are not all the

same.

Answer 1

I will use C for combination and P for permutation

a) There are C(10,3) ways to choose 3 letters out of 10

There are 26^3 ways to choose from among the letters

The numbers have at most one 9; if there are 0 9s, as we choose 7 numbers, there are 9^7 ways.

If there is one 9, there are C(7,1) = 7 ways the digit with the 9 can be selected, and 9^6 ways to choose the other 6 numbers. Thus, we have 9^6(7+9) = 16*(9^6)

C(10,3)26^3(16)(9^6)

10!/(3!7!)26^3(16)(9^6) or 10!/(3!7!)26^3(9^7+7*9^6)

17933965470720

b)There are C(10,6) ways to choose 6 letters among 10. 10!/(6!*4!)

There are 26^6 ways to choose the 6 letters.

No digit occurs twice among the 4 selected, so we have P(10,4) = 10!/6!

Thus, we have C(10,6)26^6(P(10,4)) =

10!/(6!*4!)26^6(10!/6!) or (10!)^2/((6!)^2(4!))26^6

326956457318400

c) For each possible value of digit 1, there are 999 possible values for digits 2-4 (10^3 - the 1 that all 3 values equal digit 1); for each possible value of digit 4, there are 999 possible values of digits 5-7; for each possible value of digit 7, there are 999 possible values of digits 8-10.

10*(999^3)

9970029990