n total antennas of which m are defective, confused by reasoning? Basically there are a total of n

Theresa Archer

Theresa Archer

Answered question

2022-06-17

n total antennas of which m are defective, confused by reasoning?
Basically there are a total of n antennas, of which m are defective. In how many ways can these n antennas be arranged so that no two defective antennas are side-by-side with each other?
I understand the reasoning presented in the answer to this problem in the link above. But consider the case where n = 9 and m = 7. In this case, can we say that there is no solution, because the given ( n m + 1 m ) doesn't work?

Answer & Explanation

Haggar72

Haggar72

Beginner2022-06-18Added 25 answers

The binomial coefficient will only make sense when m n m + 1, or m n + 1 2

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?