Question

if $m,n,r$ are positive integers such that $r < m,n$ then
${}^m{C_r} + {}^m{C_{r - 1}}{}^n{C_1} + {}^m{C_{r - 2}}{}^n{C_2} + .......... + {}^m{C_1}{}^n{C_{r - 1}} + {}^n{C_r}$ equals to

Answer 1

Hint:As we see these are the coefficient of ${\left( {1 + x} \right)^m} \times {\left( {1 + x} \right)^n}$ and sum of constants in base of ${}^m{C_r}$ and ${}^m{C_{r - 1}}{}^n{C_1}$ and like all other term is always $r$.Using binomial theorem we try to expand both ${\left( {1 + x} \right)^m}$,${\left( {1 + x} \right)^n}$ expressions.And then taking $x^n$ common we get an expression which is expansion of $^{m + n}{C_r}$.So using this concept we try to get the answer.

Complete step-by-step answer:
We have ${}^m{C_r} + {}^m{C_{r - 1}}{}^n{C_1} + {}^m{C_{r - 2}}{}^n{C_2} + .......... + {}^m{C_1}{}^n{C_{r - 1}} + {}^n{C_r}$
And we know that binomial theorem
${\left( {1 + x} \right)^m}$ = ${}^m{C_0} + {}^m{C_1}x + {}^m{C_2}{x^2} + ......... + {}^m{C_{m - 1}}{x^{m - 1}} + {}^m{C_m}{x^m}$
And similarly
${\left( {1 + x} \right)^n}$ = ${}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ......... + {}^n{C_{n - 1}}{x^{n - 1}} + {}^n{C_n}{x^n}$
So now we multiply both
${\left( {1 + x} \right)^m} \times {\left( {1 + x} \right)^n}$ = $({}^m{C_0} + {}^m{C_1}x + {}^m{C_2}{x^2} + ......... + {}^m{C_{m - 1}}{x^{m - 1}} + {}^m{C_m}{x^m})$ $ \times $$({}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ......... + {}^n{C_{n - 1}}{x^{n - 1}} + {}^n{C_n}{x^n})$
Now we multiply each term we get
So ${\left( {1 + x} \right)^m} \times {\left( {1 + x} \right)^n}$ = ${x^n}\left( {{}^m{C_0}{}^n{C_n} + {}^m{C_1}{}^n{C_{n - 1}} + .....} \right)$ ( we just find coefficient of ${x^n}$ there is no need to multiply whole thing )
So if we write a general term we get
${x^r}\left( {{}^m{C_0}{}^n{C_r} + {}^m{C_1}{}^n{C_{r - 1}} + .....} \right)$
As we know that ${a^m} \times {a^n} = {a^{m + n}}$
So from this ${\left( {1 + x} \right)^m} \times {\left( {1 + x} \right)^n}$ is equal to ${\left( {1 + x} \right)^{m + n}}$
So ${\left( {1 + x} \right)^{m + n}}$ = ${x^r}\left( {{}^m{C_0}{}^n{C_r} + {}^m{C_1}{}^n{C_{r - 1}} + .....} \right)$
So ${x^r}\left( {{}^m{C_0}{}^n{C_r} + {}^m{C_1}{}^n{C_{r - 1}} + .....} \right)$ is expand of $^{m + n}{C_r}$
So our required answer is $^{m + n}{C_r}$
${}^m{C_r} + {}^m{C_{r - 1}}{}^n{C_1} + {}^m{C_{r - 2}}{}^n{C_2} + .......... + {}^m{C_1}{}^n{C_{r - 1}} + {}^n{C_r}$ = $^{m + n}{C_r}$

Additional Information:Properties of binomial theorem:
1.The total number of terms in the binomial expansion of ${(a + b)^n}$ is $n + 1$, i.e. one more than the exponent $n$.
2. In the expansion, the first term is raised to the power of the binomial and in each subsequent terms the power of a reduces by one with simultaneous increase in the power of b by one, till power of b becomes equal to the power of binomial, i.e., the power of $a$ is $n$ in the first term, $\left( {n - 1} \right)$ in the second term and so on ending with zero in the last term. At the same time power of $b$ is $0$ in the first term, $1$ in the second term and $2$ in the third term and so on, ending with n in the last term.
 3. In any term the sum of the indices (exponents) of $'a'$ and $'b'$ is equal to $n$ (i.e., the power of the binomial).

Note:A simple method to solve this question :
${}^m{C_r} + {}^m{C_{r - 1}}{}^n{C_1} + {}^m{C_{r - 2}}{}^n{C_2} + .......... + {}^m{C_1}{}^n{C_{r - 1}} + {}^n{C_r}$
Now we select any one term like
${}^m{C_{r - 1}}{}^n{C_1}$ in this we add both upper power and both lower power so we get our answer
${}^m{C_{r - 1}}{}^n{C_1} = {}^{m + n}{C_{r - 1 + 1}}$
So from this we get
${}^m{C_{r - 1}}{}^n{C_1} = {}^{m + n}{C_r}$
So our required answer is ${}^{m + n}{C_r}$
[ this method is applied only when sum of upper power and lower power is always same ]