Question

Let ${T_n}$ denote the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If ${T_{n + 1}} - {T_n} = 21$ , then $n$ equals
$\left( 1 \right)$ $5$
$\left( 2 \right)$ $7$
$\left( 3 \right)$ $6$
$\left( 4 \right)$ $4$

Answer 1

We have to find the value of $n$ . We solve this using the concept of permutation and combination . We should have the knowledge of formula combinations of terms and how to expand the factorial terms . We will simplify the terms of combination in the formula to find the value of the number of sides ($n$) .

Complete step-by-step solution:
The number of terms of a combination is always one more than the combination of terms which we have to expand . If we have to find the fifth term of the combination then the value for the combination would be found for four .
Given :
${T_{n + 1}} - {T_n} = 21$
As , ${}^n{C_3}$ gives the number of triangles which can be formed using the sides of the vertices of a regular polygon of $n$ sides .
${T_n} = {}^n{C_3}$
Putting the expression of number of triangles in the given equation , we get
${}^{n + 1}{C_3} - {}^n{C_3} = 21$
We have to find the value of $n$ .
We know that , formula of combination is given as :
${}^r{C_n} = \dfrac{{r!}}{{n! \times \left( {r - n} \right)!}}$
Using the formula of combination we can write the expression as ,
$\dfrac{{\left( {n + 1} \right)!}}{{3! \times \left( {n + 1 - 3} \right)!}} - \dfrac{{n!}}{{3! \times \left( {n - 3} \right)!}} = 21$
$\dfrac{{\left( {n + 1} \right)!}}{{3! \times \left( {n - 2} \right)!}} - \dfrac{{n!}}{{3! \times \left( {n - 3} \right)!}} = 21$
On expanding the factorial , we get the expression as :
$\dfrac{{\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right)!}}{{3 \times 2 \times 1 \times \left( {n - 2} \right)!}} - \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{3 \times 2 \times 1 \times \left( {n - 3} \right)!}} = 21$
Cancelling the terms , we can write the expression as :
$\dfrac{{\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right)}}{{3 \times 2 \times 1}} - \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3 \times 2 \times 1}} = 21$
Taking $\left( n \right)\left( {n - 1} \right)$ common , we get the expression as :
$\left( n \right)\left( {n - 1} \right)\left[ {\dfrac{{\left( {n + 1} \right)}}{6} - \dfrac{{\left( {n - 2} \right)}}{6}} \right] = 21$
Cross multiplying the terms , we can write the expression as :
$\left( n \right)\left( {n - 1} \right)\left[ {n + 1 - \left( {n - 2} \right)} \right] = 21 \times 6$
$\left( n \right)\left( {n - 1} \right)\left[ {n + 1 - n + 2} \right] = 21 \times 6$
Simplifying the expression , we get
$3\left( n \right)\left( {n - 1} \right) = 21 \times 6$
Cancelling the terms , we get
$\left( n \right)\left( {n - 1} \right) = 21 \times 2$
$\left( n \right)\left( {n - 1} \right) = 42$
Splitting the value of right hand side , we can write the expression as :
$\left( n \right)\left( {n - 1} \right) = 7 \times 6$
Hence , on comparing the both sides the value of number of vertices of a regular polygon is
$n = 7$
Thus , the value of regular sides of the polygon is $7$ .
Hence , the correct option is $\left( 2 \right)$ .

Note: Corresponding to each combination of ${}^n{C_r}$ we have $r!$ permutations, because r objects in every combinations can be rearranged in $r!$ ways . Hence , the total number of permutations of n different things taken r at a time is ${}^n{C_r} \times r!$ . Thus ${}^n{P_r} = {}^n{C_r} \times r!$ , $0 < r \leqslant n$ .
Also , some formulas used :
${}^n{C_1} = n$
${}^n{C_2} = \dfrac{{n\left( {n - 1} \right)}}{2}$
${}^n{C_0} = 1$
${}^n{C_n} = 1$