Question 1201457: A security code is to be formed by using three alphabets and four digits chosen
from the alphabets {a,b,c,d,e} and digits {1,2,3,4,5,6}. All the digit and
alphabets can only be used one.Find the probability the number of different ways
of the security code can be formed if
i)all alphabets are next to each other and all digits are next to each other
ii)it consists at least two consonants
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! A security code is to be formed by using three alphabets and four digits chosen
from the alphabets {a,b,c,d,e} and digits {1,2,3,4,5,6}. All the digit and
alphabets can only be used one.Find   the number of different ways
of the security code can be formed if
i)all alphabets are next to each other and all digits are next to each other[Examples: 5126eab, dce4325]
There are P(5,3) = (5)(4)(3) = 60 ways to arrange 3 alphabet letters.
There are P(6,4) = (6)(5)(4)(3) = 360 ways to arrange 4 digits.
There are P(2,2) = 2! = 2 ways to choose whether to put the alphabet letters or
the digits first.
That's (60)(360)(2) = 43200 possible security codes for i).
ii)it consists at least two consonants.[Examples: 4d61ba2, c13d54b]
{a,b,c,d,e} contains only 3 consonants {b,c,d} and 2 vowels {a,e}.
There are two cases to choose the letters to use, exactly 2 consonants and 1
vowel or all 3 consonants. Then we'll choose the digits, then arrange the 7
chosen characters in 7! ways.
Case 1. Choose exactly 2 consonants and 1 vowel.
Choose the 2 consonants C(3,2) = 3 ways.
Choose the 1 vowel in C(2,1) = 2 ways
That's (3)(2) = 6 ways
Case 2. Choose all 3 consonants.
That's just C(3,3) = 1 way.
That's a total of 6+1 = 7 ways to choose the alphabet letters to use.
There are C(6,4) = 15 ways to choose the 4 digits.
Now we can arrange each of the (7)(15) = 105 sets of 7 characters in
P(7,7) = 7! = 5040 ways.
Answer: (105)(5040) = 529200 security codes with at least 2 consonants.
Edwin
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