Question

If N is the number of ways in which 3 distinct numbers can be selected from the set \left\\{ {{3}^{1}},{{3}^{2}},{{3}^{3}},{{........3}^{10}} \right\\} so that they form G.P. then the value of $\dfrac{N}{5}$ is.

Answer 1

Hint: We will first start by using a fact that if a, b, c are in an A.P. then ${{3}^{a}},{{3}^{b}},{{3}^{c}}$ are in G.P. Then we will use this fact to create a set of the powers of 3 from the given set. Then we will use the concept permutation and combination to find the total possible number N.

Complete step-by-step answer:
Now, before starting solution we have to understand that if a, b, c are in A. P then ${{3}^{a}},{{3}^{b}},{{3}^{c}}$ are in G.P. as we can write,
$\begin{aligned} & a=a \\\ & b=a+d \\\ & c=a+2d \\\ \end{aligned}$
Where d is a common difference.
$\begin{aligned} & \Rightarrow {{3}^{a}}={{3}^{a}} \\\ & \Rightarrow {{3}^{b}}={{3}^{a+d}} \\\ & \Rightarrow {{3}^{c}}={{3}^{a+2d}} \\\ \end{aligned}$
So, we can see that,
${{\left( {{3}^{b}} \right)}^{2}}={{3}^{a}}\times {{3}^{c}}$
$\Rightarrow {{3}^{a}},{{3}^{b}},{{3}^{c}}$ are in G.P
Now, we have a set given as \left\\{ {{3}^{1}},{{3}^{2}},{{3}^{3}},{{........3}^{10}} \right\\}.
Now, if we create a set of powers of 3 from it then we have,
\left\\{ 1,2,3,......10 \right\\}
Now, we know that if three numbers ${{3}^{a}},{{3}^{b}},{{3}^{c}}$ are in G.P. then a, b, c are in A. P. Also, we know that for three numbers a, b, c in A. P. we have,
$b=\dfrac{a+c}{2}$
So, now if a and c both are odd or both are even then the value of b will be integer as $\dfrac{a+c}{2}$ is integer if $a+c$ is even. Also, we know that for $a+c$ to be even either a, c are both odd or both even.
So, now we have to choose either two odd numbers or two even numbers from \left\\{ 1,2,3,......10 \right\\}.
So, we have the set of even numbers is \left\\{ 2,4,6,8,10 \right\\} and odd numbers is \left\\{ 1,3,5,7,9 \right\\}.
So, now we know that the number of ways of selecting r objects out n objects is,
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$
So, number of ways of selecting two even numbers from set \left\\{ 2,4,6,8,10 \right\\} is,
$\begin{aligned} & {}^{5}{{C}_{2}}=\dfrac{5!}{3!\times 2!} \\\ & =\dfrac{4\times 5}{2} \\\ & =10 \\\ \end{aligned}$
So, number of ways of selecting two odd numbers from set \left\\{ 1,3,5,7,9 \right\\} is,
$\begin{aligned} & {}^{5}{{C}_{2}}=\dfrac{5!}{3!\times 2!} \\\ & =\dfrac{4\times 5}{2} \\\ & =10 \\\ \end{aligned}$
So, the total number of ways of selecting G.P. from the set is $10+10=20=N$. Now, we have N = 20.
$\begin{aligned} & \Rightarrow \dfrac{N}{5}=\dfrac{20}{5} \\\ & \Rightarrow \dfrac{N}{5}=4 \\\ \end{aligned}$

Note: It is important to note that we have used the concept of number system that two odd numbers when added give even and the same happens for two even numbers. Using this concept we have simplified the problem to find the pair of two odd numbers or two even numbers.