How many different codes are possible?

Question

A three-digit code for certain logs uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

A. 144

B. 152

C. 160

D. 168

E. 176

Please explain your answer!

Answer 1

Hmm....this ones a doozey

You have 3 digits, each 0-9 so 10 possible for each position

BUT

1st one cannot have a zero or 1 so there are only 8 possible digits in that position

AND

2nd digit must be a zero or one so it only has 2 possible ones

Lets leave the third rules for a second and assuem the 3rd digit can be of 10 different numbers

So to get possible codes you have 3 positions with 8,2 and 10 possibles

To get this you multiply them together 8x2x10  = 160

Now, you need to eliminate to forbidden codes of 0 in 2nd and 3rd places

For each change in the first digit there is only one combination that would be forbidden (like 200, 300, 400, 500, 600, 700, 800, 900), so you just need to eliminate these 8 occurances

160-8 = 152, which is one of the answers

BTW...if there were no restrictions on digits it would be 10 x10x10...1000....what a difference a few rules make

Answer 2

The first digit cannot be 0 or 1, so there's 8 remaining digit possibilities.

The second digit must be a 0 or 1, so there's 2 digit possibilities.  But depending on what this number is, it changes the possible values in the third digit.

So we'll break this up to two problems, then add them toghether:

first:  if middle is 0:

first digit is still 8 choices

second digit is now only one choice

Since 2nd is 0, third cannot be 0, so 9 possible choices.

8 * 1 * 9 = 72

Now if middle is 1:

still 8 choices for first digit

Still 1 choice for second digit

But now all 10 choices are possible for the final digit:

8 * 1 * 10 = 80

Now add the two together:

72 + 80 = 152

B

Answer 3

152

Answer 4

B (152) - The first digit has 8 possibilities (2-9), the second digit has only 2 possibilities (0 & 1), and there are 10 possibilities for the third digit (0-9) giving you a total of 8*2*10 = 160 total code choices.  However, b/c the second and third digits cannot both be 0 that eliminates 8 codes (200, 300, 400, 500, 600, 700, 800, and 900) leaving you with 152 code choices.