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Question 5 Exercise 6.1
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====Solution==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}... n-(2 n-1))]\end{align} In the L.H.S of the above equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !... ====Solution==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n+1) !}{n !}&=\dfrac{1}{n ... n+1-(2 n))]\end{align} In the L.H.S of the above equation. are total $2 n+1$ terms \begin{align}\dfrac{(2 n
Question 3 & 4 Exercise 6.1
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on==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfra... on==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{(n+5) !}{(n+3) !}&=\dfrac{(n
Question 3 and 4 Exercise 6.2
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{r-1})$$ We are taking the right hand side of the equation \begin{align}n(^{n-1} P_{r-1})&=n \dfrac{(n-1) !}... 1} P_r+r({ }^{n-1} P_{r-1})$$ Taking R.H.S of the equation \begin{align}^{n-1} P_r+r(^{n-1} P_{r-1})&=\dfrac
Question 7 and 8 Exercise 6.2
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different arrangements can be formed of the word "equation" if all the vowels are to be kept together? ====S... olution==== The total number of alphabets in word equation are $8$, out of which $5$ are vowels. If all the
Question 2 Exercise 6.3
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0 \end{align} Let $y=n^2-3 n$ then the above last equation becomes \begin{align} & y(y+2)=840 \\ & \Rightarr