Answer :
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
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
[tex]^{n} P_{r} = \frac{n!}{(n-r)!}[/tex]
So, the total possible password = [tex]^{6}P_{r} = \frac{6!}{(6-4)!}[/tex] = [tex]\frac{6!}{4!}[/tex] = [tex]\frac{6\times5\times4!}{4!}[/tex]
On cancelling the [tex]4![/tex] in numerator and denominator we get,
Here we get, [tex]\frac{6\times5}{1}[/tex] = 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 [tex]\bold1^{st}\bold[/tex] term)
So, [tex]6\times 5 = 30[/tex]