Total numbers = 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24,
Sum of alt integers in tenth place
= 6 ( 2 + 4 + 5 + 7) = 108

\(\text { No. of lines }{ }^{10} \mathrm{C}_{2}-{ }^{4} \mathrm{C}_{2}+1=45-6+1=40\)

\({ }^{(\mathrm{n}-1)} \mathrm{C}_{\mathrm{r}}+{ }^{(\mathrm{n}+1)} \mathrm{C}_{(\mathrm{r}-1)}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\)

\({ }^{52} \mathrm{C}_{5} \text { includes all possibilities (zero king, }1 \text { king, } 2 \text { kings, } 3 \text { kings, } 4 \text { kings }){ }^{48} C_{5} \text { has no kings}\)
\(\text { Réquired possibilities }{ }^{52} \mathrm{C}_{5}-{ }^{48} \mathrm{C}_{5}\)

Number of rectangles in the chessboard is
\({ }^{9} C_{2} \times{ }^{9} C_{2} =\frac{9 \times 8}{1 \times 2} \times \frac{9 \times 8}{1 \times 2} \)
\(=36 \times 36=1296 \)

Number of 10 digit number that can be written by using the digits 2 and 3 is 2¹⁰

\(1+1 \mid 1+2\lfloor 2+3\lfloor 3+\ldots \ldots . n\lfloor n \quad=\lfloor n+1\)
\(\text { Let } \mathrm{n}=1, \quad \text { L.H.S }=1+1=2\)
\(\text { R.H.S }=\lfloor 2=2\)
It is true for n = 1, In fact it is true for n = 0 also let us assume that it is true for n=k
\(1+1 \mid 1+2\lfloor 2+3\lfloor 3+\ldots \ldots . n\lfloor n=\lfloor k+1\)
\(1+1 \mid 1+2\lfloor 2+3\lfloor 3+\ldots \ldots k \mid k+(k+1)\lfloor k+1\)
\(=\lfloor k+1+(k+1)\lfloor k+1=\lfloor k+1[1+k+1]\)
\(\lfloor k+1 (k+ 2)\)
\(\lfloor k+2\)
It is true for (k + 1)
Also by mathematical induction, it is true for all value of \(n \geq 0, n \in Z\)

\(1.3 .5 \ldots .(2 n-1) =\frac{1.2 .3 .4 \ldots \ldots .(2 n-1)(2 n)}{2.4 \ldots . \ldots(2 n)} \)
\(=\frac{\lfloor 2 n}{2^{n}\lfloor n}=\left(\frac{1}{2}\right)^{n} \cdot{ }^{2 n} C_{n} \times{ }^{n} P_{n} \)

\(\text { Given }{ }^{n} C_{4},{ }^{n} C_{5},{ }^{n} C_{6} \text { are in A.P }\)
\({ }^{2 n} \mathrm{C}_{5}={ }^{n} \mathrm{C}_{4}+{ }^{n} \mathrm{C}_{6}\)
\(\frac{2\lfloor n}{\lfloor n-5\lfloor 5}=\frac{\lfloor n}{\lfloor n -4\lfloor 4}+\frac{n}{\lfloor n -6\lfloor 6}\)
\(\frac{2}{\lfloor n-5\lfloor 5} =\frac{1}{\operatorname{\lfloor n}-4\lfloor 4}+\frac{1}{\lfloor n-6\lfloor 6}\)
\(\frac{2(n-4) 6}{(n-4)\lfloor n-5\lfloor 5.6}=\frac{5.6}{\lfloor-45.6\lfloor 4}+ \frac{(n-4)(n-5)}{\lfloor 6(n-4)(n-5) \lfloor n-6}\)
\(\Rightarrow \frac{12(n-4)}{\lfloor n-4\lfloor 6}=\frac{30}{\lfloor n-4\lfloor 6}+ \frac{(n-4)(n-5)}{\lfloor n-4 \lfloor 6}\)
\(12 n-48 =30+n^{2}-9 n+20 \)
\(n^{2}-21 n+98 =0 \)
\((n-14)(n-7) =0 \)
\(n=14(\text { or }) n =7 \)

\(1+3+5+7+\ldots \ldots+17 \text { is equal to } 9^{2}=81\)