0606 P11 - Jun 2025 - Q12 - 4 marks
7114
In this question \(n\geqslant6\).
Use an algebraic method to show that \({}^{n}\mathrm{C}_5-{}^{n-1}\mathrm{C}_5\) can be written as \({}^{n-1}\mathrm{C}_4\).
Solution
Answer: \({}^{n}\mathrm{C}_5-{}^{n-1}\mathrm{C}_5={}^{n-1}\mathrm{C}_4\).
Write the combinations in factorial form:
\({}^{n}\mathrm{C}_5-{}^{n-1}\mathrm{C}_5 =\frac{n!}{5!(n-5)!}-\frac{(n-1)!}{5!(n-6)!}.\)
Rewrite the second denominator so both terms have \((n-5)!\):
\(\frac{(n-1)!}{(n-6)!5!} =\frac{(n-1)!(n-5)}{(n-5)!5!}.\)
Also \(n!=n(n-1)!\), so
\({}^{n}\mathrm{C}_5-{}^{n-1}\mathrm{C}_5 =\frac{n(n-1)!}{5!(n-5)!} -\frac{(n-1)!(n-5)}{5!(n-5)!}.\)
Factorise:
\(=\frac{(n-1)!\{n-(n-5)\}}{5!(n-5)!}.\)
Since \(n-(n-5)=5\),
\(=\frac{5(n-1)!}{5!(n-5)!} =\frac{(n-1)!}{4!(n-5)!}.\)
But
\(\frac{(n-1)!}{4!(n-5)!}={}^{n-1}\mathrm{C}_4.\)
Therefore
\({}^{n}\mathrm{C}_5-{}^{n-1}\mathrm{C}_5={}^{n-1}\mathrm{C}_4.\)
