Let m be the number of 5-element subsets that can be chosen from the set of the first  15 natural  numbers so that at least two  of  the five numbers are consecutive. The sum of digits  of m is

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$$\begin{gathered} \mathrm{Required} \mathrm{Cases}= \\ \mathrm{Total} \mathrm{Cases}- \mathrm{Those} \mathrm{cases} \mathrm{where} \mathrm{no} \mathrm{two} \mathrm{numbers} \mathrm{are} \mathrm{consecutive} \\ {=}^{15}{\mathrm{C}}_5- \mathrm{Those} \mathrm{cases} \mathrm{where} \mathrm{no} \mathrm{two} \mathrm{numbers} \mathrm{are} \mathrm{consecutive} \\ \mathrm{For} \mathrm{those} \mathrm{cases} \mathrm{where} \mathrm{no} \mathrm{two} \mathrm{numbers} \mathrm{are} \mathrm{consecutive} \\ \mathrm{remove} 4 \mathrm{numbers} \mathrm{and} \mathrm{then} \mathrm{select} 5 \mathrm{numbers} \mathrm{which} \mathrm{covers} \mathrm{all} \mathrm{cases} \mathrm{as} \\ \mathrm{when} \mathrm{two} \mathrm{numbers} \mathrm{are} \mathrm{consecutive} \mathrm{we} \mathrm{will} \mathrm{insert} \mathrm{number} \mathrm{which} \mathrm{are} \mathrm{removed} \\ \Rightarrow \mathrm{Required} \mathrm{Cases}{=}^{\mathbf{15}}{\mathbf{C}}_{\mathbf{5}}{\mathbf{-}}^{\mathbf{11}}{\mathbf{C}}_{\mathbf{5}\mathbf{ }}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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