Question

25cn+5=25c2n-1, find n

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\sf{\underline{ \: Using \: \: the \: \: formula : \: }} \\ \\ \: \: \: \: \: \: \boxed{\sf{If \: \: \blue{ ^nC_r = ^nC_k} , then \: \: \blue{r= k }}}$

$\: \: \: \: \: \: \: \: \: \: \tt{ \pink{^{25}C_{n+5} = ^{25}C_{2n - 1}}} \\ \\ \tt{\green{ \implies n + 5 = 2n - 1 }}\\ \\ \tt {\red{ \implies n - 2n = - 1 - 5}} \\ \\ \tt{ \green{ \implies \: - n = - 6} }\\ \\ \: \: \: \: \: \: \: \: \tt{ \therefore \: \: \underline{ \pink{n = 6}}}$

$\sf{\underline{ \: Using \: \: the \: \: formula : \: }} \\ \\ \: \: \: \: \: \: \: \: \: \boxed{\sf{\: \: \blue{ \: ^nC_r = ^nC_{n-r \: } } }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \red{^{25}C_{n+5} = ^{25}C_{2n - 1}}} \\ \\ \bf{ \blue{\implies \: ^{25}C_{n+5} = ^{25}C_{25 - (2n - 1)}}} \\ \\ \bf{ \pink{\implies \: ^{25}C_{n+5} = ^{25}C_{25 - 2n + 1}}} \\ \\ \bf{ \green{\implies \: n + 5 = 25 - 2n + 1 }}\\ \\ \bf{\implies \: n + 5 = 26 - 2n} \\ \\ \bf{ \green{\implies \: n + 2n = 26 - 5 }}\\ \\ \bf{ \pink{\implies \: 3n = 21}}\\ \\ \bf{ \blue{\implies \: n = \frac{21}{3}}} \\ \\ \: \: \: \: \: \: \: \therefore \: \: \bf{ \underline{ \red{ \: n = 7 \: }}}$

Answer 2

Answer:

here is ur answer frnd

I hope it helps you