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 {{{cross(the)}}}{{{cross(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<pre>[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).</pre>

ii)it consists at least two consonants.<pre>[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</pre>