# How should I approach this combinatorics/probability problem?

My question: Consider the numbers $$1,2,3 ...,n$$. All combinations of five of these numbers are written, and one combination is chosen at random. If the probability that this chosen combination does not contain the number $$7$$ is $$0.875$$, determine the value of $$n$$.

My problem: What should I do and how should I lay this problem out (all I can think of is a simple tree diagram?)? How do I create an equation where I can make $$n$$ the subject?

The most I can do: Is it saying there are $$n! ,n! ,n! ,n! ,n!$$? So choosing $$1$$ out of $$5$$ would be $$1/5$$. In $$1/5$$, I have the probability of not containing $$7$$ $$(7/8)$$ and $$7$$ $$(1/8)$$.

• What does your concept of combination mean? Does it take order into account, allow repetition? – Mindlack Feb 20 at 12:42
• @Mindlack exactly my dood, i knew someone going to ask, but i copied down exactly how the question on my textbook is written – Fred Weasley Feb 20 at 12:48
• Isn’t there a definition in the book someplace before? – Mindlack Feb 20 at 12:51

I assume that a combination is a set of 5 distinct elements whose order does not matter. In other words, sets $$\{1,2,3,4,5\}$$ and $$\{5,4,3,2,1\}$$ represent the same combination.
The total number of combinations is $$\binom{n}{5}$$. The number of combinations with one particular number missing (say 7) is $$\binom{n-1}{5}$$.
$$\frac{\binom{n-1}5}{\binom{n}5}=\frac{\frac{(n-1)(n-2)(n-3)(n-4)(n-5)}{5!}}{\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}}=\frac{n-5}n=\frac78\implies n=40$$