Question

ncx= ncy and , x not equal to y, Then x+y=?

Answer 1

Combinations :  

$\mathsf{^{n}{C}_x\:=\:^{n} C_y}$  

Given,

x is not equal to y.  

➡️ x ≠ y   

Now,    

$\mathsf{^{n}{C}_x\:=\:^{n} C_y}$    

By using the formula,      

$\boxed{\mathsf{^{n}{C}_r\:=\:^{n} C_{(n-r) }}}$

So,  

$\mathsf{^{n}{C}_y\:=\:^{n} C_{(n-y) }}$    

$\mathsf{^{n}{C}_x\:=\:^{n} C_{(n-y)}}$    

$\mathsf{x\:=\:n\:-\:y}$    

$\mathsf{x\:+\:y\:=\:n}$      

$\boxed{\boxed{ \mathsf{x\:+\:y\:=\:n}}}$  

Combination Definition :    

Selections made by taking all or few number of objects irrelevant of its arrangement is called a combination.   

Notation :    

$\boxed{ \mathsf{^{n}{C}_r}}$.

Answer 2

$\huge{\ulcorner{\red{\sf{Answer}}}}\rfloor$

♦ Provided in question :-

→$\sf{ ^nC_x =\: ^nC_y }$

→ x ≠ y

♦ Before we start solving we must know :-

• What is Combination in Mathematics ?

→ The number of possible ways of selecting "r" number of object from given "n" number of objects .

• How to find out the Combination ?

→ Way to find out combination is

$\sf{\dfrac{n!}{(n-r)! \times r!}}$

• Representation of Combination ?

$\longrightarrow \: ^nC_r$

♦ By using above :-

$\sf{ ^nC_x = \:^nC_y }$

$\implies \sf{\dfrac{n!}{(n-x)! \times x!} = \dfrac{n!}{(n-y)! \times y!}}$

$\implies \sf{\dfrac{1}{(n-x)! \times x!} = \dfrac{1}{(n-y)! \times y!}}$

$\implies \sf{\dfrac{(n-y)!\times y!}{(n-x)! \times x!} = 1}$

♦ Then via making some conclusion :-

• That :-

$\sf{a! \times b! = b! \times a!}$

• Also x ≠ y

$\implies \sf{ (n-x)! = y! \: and \: (n-y)! = x! }$

♦ So we can say that

$\sf{(n-x) = y \: and\: (n-y) = x}$

♦ By using $\sf{(n-x) = y}$

$\implies \sf{n-x = y}$

$\implies \sf{ n = x + y }$

Or

$\large{\boxed{\boxed{\sf{x+y=n}}}}$