# $4$ integers are randomly selected from the numbers from $1$ to $10$. The chance that there are at least two successive numbers among those $4$ is

$4$ integers are randomly selected from the numbers from $1$ to $10$. The chance that there are atleast two successive numbers among those $4$ selected is
$(A)\frac{5}{6}\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{2}{3}\hspace{1cm}(D)\frac{1}{2}\hspace{1cm}$

I calculated answer as $\frac{24}{\binom{10}{4}}$ but this wrong. Please help me find the right answer.

• Hint: it is much easier to calculate the complementary probability that all 4 numbers are spaced apart by at least 1. – Erick Wong Sep 29 '15 at 5:18
• Please elaborate i am not getting the answer. @ErickWong – user1442 Sep 29 '15 at 6:44
• Is the selection with or without replacement? – robjohn Sep 29 '15 at 10:03
• See 'math.stackexchange.com/questions/1295252' for a closely related, slightly simpler, problem. – BruceET Feb 3 '16 at 0:10

Take out $4$ numbers, 6 numbers (N) are left with $7$ possible gaps including ends

$_ N _ N _ N _ N _ N _ N _$

We can replace the $4$ numbers in forbidden way in any of 7 gaps (including ends) in $\dbinom{7}{4}$ ways

thus indicated $Pr = 1 - \dfrac{\dbinom{7}{4}}{\dbinom{10}{4}}$

Here is a simulation in R with 100,000 performances of the experiment (choices without replacement), which verifies the Answer $5/6$ of @trueblueanil within simulation error.

 m = 10^5;  d = numeric(m)
for(i in 1:m) {
x = sample(1:10, 4);  sort(x)
d[i] = min(diff(sort(x)))  }
mean(d == 1)  # 'd==1' is logical vector of T's and F's
## 0.83222    #  its 'mean' is proportion of T's
1 - choose(7,4)/choose(10,4)  # exact analytic answer
## 0.8333333
2*sqrt((.83)*(.17)/m)  # 95% margin of simulation error
## 0.002375710


To get greater accuracy, use m=10^6, but run shown is sufficient to distinguish among the four answers proposed.

I came up with the following reasoning :

Pattern 1: At least two successive numbers, \eqalign { & (**)(*)(*) \cr & (*)(**)(*) \cr & (*)(*)(**) \cr & choices \ = 3 \times (2!\cdot 9) \times 8 \times 7 = 3024 } Pattern 2: At least three successive numbers, \eqalign { & (***)(*) \cr & (*)(***) \cr & choices \ = 2 \times (3!\cdot 8) \times 7 = 672 }

Pattern 3: At least four successive numbers, \eqalign { & (****) \cr & choices \ = 1 \times (4!\cdot 7) \times 6 = 1008 } The total number of legal possible choices, using the inclusion exclusion principle is : $$3024-672+1008 = 3360$$ and the required probability is $${3360 \over 10 \cdot 9 \cdot 8 \cdot 7} = {2 \over 3}$$