How many different sequences of 4 digits are possible if the first digit must be 3, 4, or 5 and if the sequence may not end in 000? Repetition of digits is allowed.
A. 1,512

B. 2,999

C. 2,997

D. 2,000

Question

How many different sequences of 4 digits are possible if the first digit must be 3, 4, or 5 and if the sequence may not end in 000? Repetition of digits is allowed. 
 A. 1,512  
 
 B. 2,999  
 
 C. 2,997  
 
 D. 2,000

kinggeorge · Answer

You have three choices for the first digit. As for the last three digits, it can't be 000. This means that it can 0, 1, or 2 zeros. just not 3. So let's find the total number of ways, and then subtract the ones that have 000 for the end.

anonymous · Answer

alrighhty

kinggeorge · Answer

The total number of ways would be given by $$3\cdot\left( 10^3\right)$$since there are 10 options for for the last three digits. 

Then, the number of ways that we can end in 000 is just 3. It has to be either 3000, 4000, or 5000.

kinggeorge · Answer

Hence, there are $$3\cdot10^3 -3$$ways to make this number.

anonymous · Answer

which would be what answer above ? a,b,c,d ?

kinggeorge · Answer

$$3\cdot10^3-3=3\cdot1000-3=3000-3=2997$$