Combinatorics and Probability Problem The problem I am working on is:
An ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8, or 9, in succession.
a.How many different possible PINs are there if there are no restrictions on the choice of digits?
b.According to a representative at the author’s local branch
of Chase Bank, there are in fact restrictions on the choice
of digits. The following choices are prohibited: (i) all four
digits identical (ii) sequences of consecutive ascending or
descending digits, such as 6543 (iii) any sequence start-ing with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the prob-ability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?
c. Someone has stolen an ATM card and knows that the first
and last digits of the PIN are 8 and 1, respectively. He has
three tries before the card is retained by the ATM (but
does not realize that). So he randomly selects the $2nd$ and $3^{rd}$
digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the
individual knows about the restrictions described in (b)
so selects only from the legitimate possibilities). What is
the probability that the individual gains access to the
account?
d.Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively. 
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For part a): The total number of pins without restrictions is $10,000$
For part b): The number of pins in either ascending or descending order is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known, then the three other spots containing digits are already spoken for. The number of pins where each slot contains the same digit is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known there is only one option left to the rest of the slots. The number of pins that have their first and second slot occupied by 1 and 9, respectively, is $1 \cdot 1 \cdot 10 \cdot 10 \cdot$. So, if R is the set that contains these restricted pins, then $|R| = 130$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N| = 10,000 - 130$. Hence, the probability is then $P(N) = 9780/10000 = 0.9870.$ However, the answer is $0.9876$. What did I do wrong?
For part c): The sample space, containing all of the outcomes of the experiment that will take place, is $|N|=9870$. When it says that the thief won't use the same pair of digits in each try, does that not allow him trying the pin 8 5 2 1 in one try and the pin 8 2 5 1 in another try?
 A: For b): Which is the descending sequence starting with $1$ that you counted?
For c): Good question; the problem is badly worded in that regard. Taking it literally, I'd tend to interpret it as referring to unordered pairs, but since it makes little sense to couple two different PINs in this manner, I suspect that they actually mean ordered pairs. However, note that the answer doesn't depend on this.
I understand neither why the question says that the thief knows the restrictions, nor why you say that the sample space has size $9870$. The thief knows that the first and last digits are $8$ and $1$, respectively; that's not compatible with any of the sequences excluded by the restrictions, and it doesn't allow for $9870$ possibilities.
A: For d): We have the case: $1$ * * $1$. 
But the thief knows, by prohibition (i), it can not be $1111$. Thus he eleminates $1$ possibility.
Also he knows, by (iii), the second digit can not be $9$. There are exactly $10$ number of the form $19$ * $1$, namely, $1900$, $1910$, $1920$... So, at this stage he eleminates $10$ possibilities.
All in all, if he had no restrictions, there were $100$ choices for the form $1$ * * $1$. But he excluded $10 + 1 = 11$ of them and get 89 possible choices. Since he has 3 chances, the resulting possibility is $3/89$.
A: For (c) he has 3 tries, and there are a total of 100 choices. There are 10 choices for the 2nd number and 10 choices for the 3rd number. Since the PIN begins with an 8 and ends with a 1, none of the restrictions apply.

*

*For his first try, there is a $1/100 = 0.0100$ probability of guessing the correct PIN.

*For his second try, there is a $1/99 \approx 0.0101$ probability.

*For his third try, there is a $1/98 \approx 0.0102$ probability.

So the total probability is $1/100 + 1/99 + 1/98 \approx 0.0303$.
A: Just my two cents, 
10^4 possibilities.  
There are 14 ascending and descending groups of 4.
Keyspace 9,876
If the badguy knows two of the four spaces, he only has to guess through entropy 10^2.
(None of the restrictions meet up with the range 8xx1)
3 Tries in 100
:)
A: Regarding part [c], I do not think that there is something wrong with the book. As @Mack said in one of his comments, thief has to guess only two second numbers. Since the given restrictions actually do not apply on this order of numbers, he has to guess among 100 (10*10) possible combinations of the numbers.
Since he has 3 tries, the probability that he will gain desired access is 3/100.
According to Wolfram Alpha, 3/100 is exactly 0.03, which is the same as written solution from the book.
Correct me if I am wrong.
