3 cards are randomly selected from a desk of 9 cards uniquely number from 1 to 9. What is the probability that no two of the cards selected have numbers that differ by 1? Click here to show the solutions.
Solution 1 Let’s denote $f(m)$ as total number of 3-card groups that can be selected with $m$ as the smallest number in each of the groups, and $g(m,n)$ as the total number of 3-card groups that can be selected with $m$ as the smallest number and $n$ as the second smallest number. Each group meets the requirement that no two numbers differ by 1.
Therefore, we have
$$f(1)=\sum_{i=3}^{7}g(1,i)=5+4+3+2+1=15$$ $$f(2)=\sum_{i=4}^{7}g(2,i)=4+3+2+1=10$$ $$f(3)=\sum_{i=5}^{7}g(3,i)=3+2+1=6$$ $$f(4)=\sum_{i=6}^{7}g(4,i)=2+1=3$$ $$f(5)=\sum_{i=7}^{7}g(5,i)=1$$ $$f(6)=f(7)=f(8)=f(9)=0$$
Therefore, the total number of 3-card groups that meet the restriction is $$\sum_{i=1}^{9}f(i)=15+10+6+3+1+0+0+0+0=35$$ Therefore the answer to the question is $$\dfrac{35}{C(9,3)}=\dfrac{35}{\dfrac{9\times 8\times 7}{3\times 2\times 1}}=\dfrac{5}{12}$$
Solition 2 Let’s consider 3-card groups that do not meet the requirement. There are two cases: one with only one pair of numbers differing by 1, such as (1, 2, 4); the other case with two pairs of numbers differing by 1, such as (2, 3, 4).
For the first case, denote $h(x, y)$ is the number of 3-groups with only pair of numbers differing by $1$, $y = x + 1$ . Therefore the total number of 3-card groups with only one pair numbers differing by 1 is $$h(1,2)+h(2,3)+h(3,4)+h(4,5)+h(5,6)+h(6,7)+h(7,8)+h(8,9)$$ $$=6+5+5+5+5+5+5+6=42$$
For the second case, there are 7 3-card groups: $(1,2,3), (2,3,4), …, (7,8,9)$. Therefore, the total number of 3-card groups that do not meet the restriction is $42 + 7 = 49$.
Therefore the answer to the question is $$1-\dfrac{49}{C(9,3)}=1-\dfrac{49}{\dfrac{9\times 8\times 7}{3\times 2\times 1}}=\dfrac{5}{12}$$