Question

Find r,if 11c4+11c5+12c6+13c7=14cr

Answer 1

$\textbf{Concept used:}$

$\bf\,^nC_r+^nC_{r-1}=^{(n+1)}C_r$

$\text{Given:}$

$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7=^{14}C_r$

$^{11}C_5+^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7=^{14}C_r$

$\implies^{12}C_5+^{12}C_6+^{13}C_7=^{14}C_r$

$\implies^{12}C_6+^{12}C_5+^{13}C_7=^{14}C_r$

$\implies^{13}C_6+^{13}C_7=^{14}C_r$

$\implies^{13}C_7+^{13}C_6=^{14}C_r$

$\implies^{14}C_7=^{14}C_r$

$\text{Using}$

$\boxed{\bf^nC_x=^nC_y\implies\;x=y\;\;(or)\;\;n=x+y}$

$\implies\bf\,r=7$

Answer 2

r=7

Step-by-step explanation:

$(11C_4+11C_5)+12C_6+13C_7=14C_r$

We know that

$nC_r+nC_{r-1}=(n+1)C_r$

n=11 and r=5

Using this formula

$12C_5+12C_6+13C_7=14C_r$

n=12 and r=6

Again using  the formula

$(12+1)C_6+13C_7=14C_r$

$13C_6+13C_7=14C_r$

n=13, r=7

Again using the formula

$(13+1)C_7=14C_r/tex]</p><p>[tex]14C_7=14C_r$

By comparing on both sides then we get r=7

#Learns more:

https://brainly.in/question/13505350