#### Eagerissac

##### New member

- Joined
- Jan 9, 2020

- Messages
- 6

GIVEN QUESTION:

A decimaI-string is a sequence of characters consisting of {0,1,2,3,4,5,6,7,8,9}.

Etc:

0

3463412398

Let n≥0 be an integer.

1. Calculate the number of decimaI-strings of Iength n.

So I know the answer to this is 10^n

2. Calculate the number of decimal-strings d1,…,dn of length n such that d1d2≠00

*and*d2d3≠01?

Following my teacher's example below, I counted 981 valid decimal-strings when n = 3. However, when I tried to test the formula I got for n = 4, I counted 9801 decimal strings when the formula should give me 9810 valid decimal-strings. I re-counted the valid strings from each length multiple times, but I can't get the formula to work for n>= 4. I was wondering if someone could explain what I'm doing wrong?

981 x 10^(n-3)

3. Calculate the number of decimal-strings d1,…,dn of length n such that d1d2=00

*or*d1d2d3=111

This one I'm a little confused as to how the "or" would change my calculations and was hoping for some help.

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TEACHER'S SOLUTION FOR SIMILAR QUESTION THAT I WAS BASING MY WORK OFF OF:

An n-bit

*binary string*is a sequence of length n over the alphabet {0,1}.

1. How many n-bit binary strings are there?

2^n

2. How many n-bit binary strings b1,…,bn are there such that b1b2≠00 and such that b2b3≠01?

These are the valid choices of b1b2b3: {011, 010, 100, 110, 111}. (The invalid choices are {000, 001, 101}.) After choosing b1b2b3 we have 2 choices for each of b4, . . . , bn. Therefore, by the Product Rule, the number of strings we are interested in is

5 × 2^(n−3)