Y={10}

\(Step\quad I\quad Let\quad P(n)\quad be\quad the\quad given\quad statement. i.e.P(n):{ 7 }^{ n }-{ 2 }^{ n }is\quad divisible\quad by\quad 5.\)
\(Step\quad II\quad For\quad n=1,\quad we\quad have\quad P(1):{ 7 }^{ 1 }-{ 2 }^{ 1 }=5\)
Which is divisible by 5.
Thus,P(1) is true.
\(Step\quad III\quad Let\quad us\quad assume\quad that\quad P(n)\quad is\quad true\quad for\quad n=k. i.e.P(k):{ 7 }^{ k }-{ 2 }^{ k }is\quad divisible\quad by\quad 5.\)
\(Then\quad { 7 }^{ k }-{ 2 }^{ k }\quad =5d\quad for\quad some\quad d\in N.\)
\(\Rightarrow { 7 }^{ k }=5d+{ 2 }^{ k },for\quad some\quad d\in N.\quad ...(i)\)
Step IV Now, we shall prove the statement for n=k+1.
For this, we have to show that N.
\({ 7 }^{ k+1 }-{ 2 }^{ k+1 }is\quad divisible\quad by\quad 5.\)
\(Then,\quad { 7 }^{ k+1 }-{ 2 }^{ k+1 }={ 7 }^{ k }.7-{ 2 }^{ k+1 }\)
\(=7(5d+{ 2 }^{ k })-{ 2.2 }^{ k }\quad \quad [from\quad Eq.(i)]\)
\(=35d+7.2k-{ 2.2 }^{ k }=35d+{ 5.2 }^{ k }\)
\(=5(7d+{ 2 }^{ k }),which\quad s\quad divisible\quad by\quad 5.\)

Thus,P(k+1) is true, when ever P(k) is true.Hence,by principle of mathematical induction,P(n) is true for all n∈N.

Step I Let P(n) be the given statement.i.e., P(n) :\(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ n-1 }\theta =\frac { { sin2 }^{ n }\theta }{ { 2 }^{ n }sin\theta } \)
Step II For n = 1, we have, LHS = \(cos\theta\) and \(RHS=\frac { sin2\theta }{ { 2 }sin\theta } =\frac { { 2 }sin\theta cos\theta }{ { 2 }sin\theta } =cos\theta \)
\(\therefore\) LHS = RHS  P (1) is true.
Step III Let us assume that P(k) is true. i.e., P(k) : \(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta =\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } \ ...(i)\)
Step IV Now, we shall prove the statement for n = k + 1
For this, we have to show that
\({ cos2 }^{ 2 }\theta ...{ cos2 }^{ (k+1)-1 }\theta \quad \frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta }\)
\(Then \ LHS=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k }\theta \)
\(=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta \ { cos2 }^{ k }\theta\)
\(=\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } .{ cos2 }^{ k }\theta\)
\([Multiplying \ numerator \ and \ denominator \ by \ 2]\)
\(=\frac { { sin2.(2 }^{ k }\theta ) }{ { 2 }^{ k+1 }sin\theta } \ \ [\because 2sin\theta \ cos\theta =sin2\theta ]\)
\(=\frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta } =RHS\)
\(Thus, \ P(k+1) \ is \ true, \ whenever \ P(k) \ is \ true.\)
\(Hence, \ by \ principle \ of \ mathematical \ induction, \ P(n) \ is \ true \ for \ all \ n\in N\)

Let the two sets be A and B such that n(A) =m and n(B) =n.
Then, number of subsets of set A=2^m
and the number of the subset of set B=2ⁿ
According to given condition, we have
2^m=112+2ⁿ\(\Rightarrow\)2^m-2ⁿ=2⁷-2⁴
On comparing both sides, we get
2^m=2⁷ and 2ⁿ = 2⁴ \(\Rightarrow\)m=7 and n= 4

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)