How do you solve this probability problem? Tad draws three cards at random, without replacement, from a deck of ten cards numbered 1 through 10. What is the probability that no two of the cards drawn have numbers that differ by 1? Express your answer as a common fraction.
 A: Encode the possible choices as $\{0,1\}$-words of length $10$ containing three ones. There are ${10\choose3}$ such words, all of them equiprobable. In order to count the admissible words choose any word of length $8$ containing three ones, and insert a zero after the first and the second one. It follows that the probability $p$ we are after amounts to 
$$p={{8\choose3}\over{10\choose3}}={8\cdot7\cdot 6\over 10\cdot 9\cdot 8}={7\over15}\ .$$
A: Take the complementary cases, where at least two cards are differing by $1$, and all three cards are differing by $1$. Now, all three cards are differing by $1$, it can be happened in $8$ ways $(1,2,3 ; 2,3,4 ; ... 8,9,10)$. Now check the cases where two cards are differing by $1$. For the end pairs $(i.e., 1,2$ and $9,10)$ , you'll get $7$ cases for each. And for the other pairs, you will get $6$ cases for each. Like I say, take the pair $(4,5)$. $3$ can't be counted as for all three cards are differing by $1$, this was considered as $(3,4,5)$. By similar argument, we can't take $6$. Hence only $6$ cards can be taken, as it's for without replacement, you can't take $4$ and $5$. Hence, outcome $=$ $8 + 7\times 2 + 6\times 7$ $=$ $64$. Hence the required probability is $=$ $1$ $-$ ${64\over 120}$ , i.e., ${7\over 15}$
A: HINT: Consider the complement; how many ways are there to choose three cards in such a way that two cards have numbers that differ by one?
