====== Question 4 Exercise 6.3 ======
Solutions of Question 4 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 4(i)=====
Prove that: ${ }^{n-1} C_r+{ }^{n-1} C_{r-1}={ }^n C_r$
====Solution====
Given that:
$${ }^n{ }^1 C_r+{ }^n{ }^1 C_{r-1}={ }^n C_s$$
Taking L.H.S
\begin{align}
{ }^{n-1} C_r+{ }^{n-1} C_{r-1}&=\dfrac{(n-1) !}{(n-r-1) ! r !}+\dfrac{(n-1) !}{(n-1-(r-1)) !(r-1) !} \\
& =\dfrac{(n-1) !}{(n-r-1) ! r(r-1) !}+\dfrac{(n-1) !}{(n-r) !(r-1) !} \\
& =\dfrac{(n-1) !}{(n-r-1) ! r(r-1) !}+\dfrac{(n-1) !}{(n-r)(n-r-1) !(r-1) !} \\
& =\dfrac{(n-1) !}{(n-r-1) !(r-1) !}[\dfrac{1}{r}+\dfrac{1}{n-r}] \\
& =\dfrac{(n-1) !}{(n-r-1) !(r-1) !} \cdot[\dfrac{n-r+r}{r(n-r)}] \\
& =\dfrac{n(n-1) !}{(n-r)(n-r-1) ! r(r-1) !} \\
& =\dfrac{n !}{(n-r) ! r !} \\
& ={ }^n C_r=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}\end{align}



=====Question 4(ii)=====
Prove that: $r \cdot{ }^n C_r=n^{n-1} C_{r-1}$
====Solution====
Given that:
$$r^n C_r=n^{n-1} C_{r-1} $$
We take L.H.S
\begin{align}
 r^n C_r&=r \dfrac{n !}{(n-r) ! r !} \\
 \Rightarrow r^n C_r&=r \dfrac{n(n-1) !}{(n-1-(r-1)) r(r-1) !}\\
r .{ }^n C_r&=n \cdot \dfrac{(n-1) !}{(n-1-(r-1))(r-1) !} \\
& =n^{n-1} C_{r-1}\end{align}






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