15) All possible 120 permutations of WDSMC are arranged in dictionary order, as if each were an ordinary five -  letter word. The last letter of the 86^th  word in the list is :
(a) W
(b) D
(c) M
(d) C

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{WDSMC} \\ \mathrm{CDMSW} \\ \mathrm{Words} \mathrm{which} \mathrm{start} \mathrm{with} \mathrm{C}\to \mathrm{Arrangement} \mathrm{of} \mathrm{remaining} \mathrm{letters}=4!=24 \\ \mathrm{Similarly} \\ \mathrm{Words} \mathrm{which} \mathrm{start} \mathrm{with} \mathrm{D}=4!=24 \\ \mathrm{Words} \mathrm{which} \mathrm{start} \mathrm{with} \mathrm{M}=4!=24 \\ \mathrm{First} \mathrm{word} \mathrm{with} \mathrm{S} \mathrm{is} \mathrm{SCDMW} \\ \mathrm{Words} \mathrm{which} \mathrm{start} \mathrm{with} \mathrm{SC}=3!=6 \\ \mathrm{Words} \mathrm{which} \mathrm{start} \mathrm{wiht} \mathrm{SD}=3!=6 \\ \mathrm{Words} \mathrm{which} \mathrm{start} \mathrm{with} \mathrm{SMC}=2!=2 \mathrm{which} \mathrm{SMCDW} \mathrm{and} \mathrm{SMCWD} \\ 24+24+24+6+6+2=86 \\ \mathrm{Hence} 86\mathrm{th} \mathrm{word} \mathrm{si} \mathrm{SMCWD} \\ \mathrm{Hence} \mathrm{option}\left(\mathrm{B}\right) \mathrm{is} \mathrm{correct} \end{gathered}$$

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