If { }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ } | vedclass

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(C) We use the Pascal's identity: ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$.Given expression: ${ }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ }^{13} C

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) We use the Pascal's identity: ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$.Given expression: ${ }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ }^{13} C_7={ }^{14} C_{r}$.Applying the identity to the first two terms: ${ }^{11} C_4+{ }^{11} C_5 = { }^{12} C_5$.Now the expression becomes: ${ }^{12} C_5+{ }^{12} C_6+{ }^{13} C_7$.Applying the identity again: ${ }^{12} C_5+{ }^{12} C_6 = { }^{13} C_6$.Now the expression becomes: ${ }^{13} C_6+{ }^{13} C_7$.Applying the identity one last time: ${ }^{13} C_6+{ }^{13} C_7 = { }^{14} C_7$.Comparing this with ${ }^{14} C_{r}$,we get $r = 7$.

If ${ }^{11} C_4+{ }^{11} C_5+{ }^{12} C_6+{ }^{13} C_7={ }^{14} C_{r}$,then the value of $r$ is

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