A committee of 4 is to be chosen from a group of students. If the number of students in the group increases by 1, the number of different committees triples. How many students are in the group?
So we have this equation: \(3{ n \choose 4} = {n + 1 \choose 4}\)
Simplify to: \(3 \times { n! \over {n -4}! \times 24} = { n+1! \over {n -3}! \times 24} \)
Now, further simplify: \({n! \over (n-4)! \times 8} = {(n+1)! \over (n-3)! \times 24}\)
Now, note that \(n! = (n-4)! \times (n-3) \times (n-2) \times (n-1) \times n\).
Applying this logic to both sides gives us \({(n -3)(n-2)(n-1)n\over 8} = {n(n +1)(n-1)(n-2)\over 24}\)
Cross multiplying gives us \({(n -3)(n-2)(n-1)24n} = {8n(n +1)(n-1)(n-2)}\)
Canceling out like terms gives us \(8(n+1) = 24(n-3)\), meaning \(n = \color{brown}\boxed5\)