Question

Amy is choosing a 2-letter password from the letters A, B, C, D, E, and F. The password cannot have the same letter repeated in it. How many such passwords are possible?

Answer 1

Answer:

Amy can choose from 60 possible passwords.

Solution:

Given that the two-letter password is from the letters A, B, C, D, E and F

Total number of letters = 6  

The password is a two digit letter and repetition of letters is not allowed.

Method 1:

\therefore The number of ways of obtaining an ordered subset of r elements from a set of n elements is given by

$^{n} P_{r} = \frac{n!}{(n-r)!}$

So, the total possible password = $^{6}P_{r} = \frac{6!}{(6-4)!}$ = $\frac{6!}{4!}$ = $\frac{6\times5\times4!}{4!}$

On cancelling the $4!$ in numerator and denominator we get,

Here we get, $\frac{6\times5}{1}$ = 30  possible passwords

Method 2:

There are 6 possible choices for the first term of the password and 5 choices for the second term (i.e. leaving the selected letter of the $\bold1^{st}\bold$ term)

So, $6\times 5 = 30$

Answer 2

Answer:

30

Step-by-step explanation:

6 choices for the 1st, then 5 choices for the 2nd, so: 6*5=30

or  nPk=n!/(n-k)!  with n=6 and k=2

Put it simply each letter can be combined with 5 diffrent letters and with 6 total letters 6x5=30