Question

Find r, if ^16C4+^C5+^17C6+^18C7=^19Cr
​

Answer 1

Appropriate Question :-

Find r, if

$\qquad\sf \: \ ^{16}C_4 + \ ^{16}C_5 + \ ^{17}C_6 + \ ^{18}C_7 = \ ^{19}C_r$

Answer:

$\qquad\qquad \qquad\boxed{ \sf{ \:\bf \: r = 7 \:, \: 12 \: }}$

Step-by-step explanation:

Given that,

$\qquad\sf \: \ ^{16}C_4 + \ ^{16}C_5 + \ ^{17}C_6 + \ ^{18}C_7 = \ ^{19}C_r$

$\qquad\sf \: \: (\ ^{16}C_4 + \ ^{16}C_5) + \ ^{17}C_6 + \ ^{18}C_7 = \ ^{19}C_r$

We know,

$\qquad\boxed{ \sf{ \:\ ^{n}C_r \: + \: \ ^{n}C_{r - 1} = \ ^{n + 1}C_r \: }}$

So, using this result, we get

$\qquad\sf \: \: \ ^{17}C_5 + \ ^{17}C_6 + \ ^{18}C_7 = \ ^{19}C_r$

$\qquad\sf \: \: (\ ^{17}C_5 + \ ^{17}C_6) + \ ^{18}C_7 = \ ^{19}C_r$

Again, using the same result, we get

$\qquad\sf \: \: \ ^{18}C_6 + \ ^{18}C_7 = \ ^{19}C_r$

$\qquad\sf \: \: \ ^{19}C_7 = \ ^{19}C_r$

We know,

$\qquad\boxed{ \sf{ \:\begin{gathered}\begin{gathered}\bf\: \ ^{n}C_x = \ ^{n}C_y \: \sf\implies \begin{cases} &\sf{x = y}\\ &\sf \: or \\ &\sf{n = x + y} \end{cases}\end{gathered}\end{gathered} \: }}$

So, using this result, we get

$\qquad\sf \: r = 7 \: \: or \: \: 19 = r + 7$

$\qquad\sf\implies \bf \: r = 7 \: \: or \: \: r = 12$