Question

If \[\dfrac{1}{^{4}{{C}_{n}}}=\dfrac{1}{^{5}{{C}_{n}}}+\dfrac{1}{^{6}{{C}_{n}}}\], then n =

Answer 1

Hint: The given question is based on combinations. We will be using the formula of combinations to solve for ‘n’. The formula we have is, \[^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}\], where ‘n’ is total possible arrangements possible and ‘r’ is the arrangements selected. Applying this formula in the given expression, we will expand the expression a little and cancel out the similar terms and then we will solve further to get the value of ‘n’.

Complete step by step answer:
According to the given question, we have questions based on combinations formulas. The expression given to us has to be solved for ‘n’.
The formula of combinations we have is,
\[^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}\] where ‘n’ is total possible arrangements possible and ‘r’ is the arrangements selected.
So, we will have to apply this formula in the given expression to find the value of ‘n’.
The expression we have is,
\[\dfrac{1}{^{4}{{C}_{n}}}=\dfrac{1}{^{5}{{C}_{n}}}+\dfrac{1}{^{6}{{C}_{n}}}\]
Applying the formula, we get,
\[\Rightarrow \dfrac{1}{\dfrac{4!}{n!(4-n)!}}=\dfrac{1}{\dfrac{5!}{n!(5-n)!}}+\dfrac{1}{\dfrac{6!}{n!(6-n)!}}\]
\[\Rightarrow \dfrac{(4-n)!}{4!}=\dfrac{(5-n)!}{5!}+\dfrac{(6-n)!}{6!}\]
Opening up the factorials and cancelling up the similar terms, we get,
\[\Rightarrow \dfrac{(4-n)!}{4!}=\dfrac{(5-n)(4-n)!}{5!}+\dfrac{(6-n)(5-n)(4-n)!}{6!}\]
Cancelling up the common terms across the equality, we get
\[\Rightarrow \dfrac{1}{4!}=\dfrac{(5-n)}{5!}+\dfrac{(6-n)(5-n)}{6!}\]
\[\Rightarrow \dfrac{1}{4!}=\dfrac{(5-n)}{5\times 4!}+\dfrac{(6-n)(5-n)}{6\times 5\times 4!}\]
\[\Rightarrow \dfrac{1}{1}=\dfrac{(5-n)}{5}+\dfrac{(6-n)(5-n)}{6\times 5}\]
Taking the common terms out from the above equation, we get,
\[\Rightarrow (5-n)\left[ \dfrac{1}{5}+\dfrac{(6-n)}{30} \right]=1\]
Solving the left over expression, we get,
\[\Rightarrow (5-n)\left[ \dfrac{12-n}{30} \right]=1\]
\[\Rightarrow (5-n)(12-n)=30\]
Opening up the brackets now, we will get a quadratic equation and we will solve further to get the value of ‘n’.
 We have,
 \[\Rightarrow {{n}^{2}}-17n+30=0\]
Solving the quadratic equation, we get,
\[\Rightarrow {{n}^{2}}-15n-2n+30=0\]
\[\Rightarrow n(n-15)-2(n-15)=0\]
Taking the common terms out and then equating it to zeroes, we will get,
\[\Rightarrow (n-15)(n-2)=0\]
\[n=15,2\]
So, ‘n’ has two values, but we know that ‘n’ cannot be more than 4. So we will have \[n=2\].

Therefore, the value of \[n=2\].

Note: While calculating the equations, either solve it separately or do it step wise, so that there are no mistakes. The value of ‘n’ has to be less than 4 as from the given expression we can see clearly that ‘n’ has to satisfy for all the combinations given, so the least value will be suitable for all.