If { }^{15} C_4+{ }^{15} C_5+{ }^{16} C_6+{ } | vedclass

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(D) We use the identity ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$.First,consider the first two terms: ${ }^{15} C_4+{ }^{15} C_5 = { }^{16} C_5$

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$8$ or $11$

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(D) We use the identity ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$.First,consider the first two terms: ${ }^{15} C_4+{ }^{15} C_5 = { }^{16} C_5$.Now,the expression becomes ${ }^{16} C_5+{ }^{16} C_6+{ }^{17} C_7+{ }^{18} C_8$.Next,${ }^{16} C_5+{ }^{16} C_6 = { }^{17} C_6$.Now,the expression becomes ${ }^{17} C_6+{ }^{17} C_7+{ }^{18} C_8$.Next,${ }^{17} C_6+{ }^{17} C_7 = { }^{18} C_7$.Now,the expression becomes ${ }^{18} C_7+{ }^{18} C_8$.Finally,${ }^{18} C_7+{ }^{18} C_8 = { }^{19} C_8$.Given ${ }^{19} C_8 = { }^{19} C_{r}$,we know that ${ }^{n} C_{x} = { }^{n} C_{y}$ implies $x = y$ or $x + y = n$.Here,$r = 8$ or $r = 19 - 8 = 11$.

If ${ }^{15} C_4+{ }^{15} C_5+{ }^{16} C_6+{ }^{17} C_7+{ }^{18} C_8={ }^{19} C_{r}$,then the value of $r$ is equal to

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