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?
+2666 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 \(n = \color{brown}\boxed5\)
