Question 1206249: " Can you please help me with this question by explaining the steps, "step by step"? I am stuck and just cant get the idea on how to do it. The problem is, " HOW MANY DIFFERENT LICENSE PLATES OF 5 DIGITS ARE POSSIBLE IF THE FIRST DIGIT MUST BE A NUMBER, THE NEXT THREE DIGITS ARE LETTERS, AND THE LAST DIGIT IS A NUMBER.
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Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there are 10 numbers you can choose from.
they are 0,1,2,3,4,5,6,7,8,9.
there are 26 letters you can choose from.
they are a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z
assuming all letters and numbers can be entered more than once, your possible number of entries would be.
first didit = 10
second digit = 26
third digit = 26
fourth digit = 26
fifth digit = 10
total number of possible combination would be 10 * 26 * 26 * 26 * 10 = 10^2 * 26^3 = 1,757,600 possible choices.
to see how this works, make the possible choices less.
allow for 2 numbers per digit and 2 letters per digit.
let the numbers = 1,2
let the letters = a,b
the possible choices would be 2^2 * 2^3 = 4 * 8 = 32
those choices would be:
1AAA1 1AAA2 1AAB1 1AAB2 1ABA1 1ABA2 1ABB1 1ABB2 1BAA1 1BAA2 1BAB1 1BAB2 1BBA1 1BBA2 1BBB1 1BBB2
16 MORE COMBINATIONS WITH 2 AS THE FIRST DIGIT.
TOTAL = 32.
it works the same way with 10 * 26 * 26 * 26 * 10 = 10^2 * 26^3.
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