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Question Number 111906 by mr W last updated on 05/Sep/20

Answered by PRITHWISH SEN 2 last updated on 22/Sep/20

$$\mathrm{Disarrangement}\mathrm{of}6\mathrm{cards}!6=265$$$$\mathrm{now}\mathrm{in}\mathrm{this}265\mathrm{ways}\mathrm{card}\mathrm{no}.1\mathrm{can}\mathrm{be}\mathrm{in}$$$$\mathrm{envelope}\mathrm{no}.2,3,4,5\mathrm{or}6\mathrm{and}\mathrm{in}\mathrm{each}\mathrm{case}\mathrm{the}$$$$\mathrm{no}.\mathrm{of}\mathrm{ways}\mathrm{will}\mathrm{be}\mathrm{same}.$$$$\therefore \mathbf{The}\mathbf{required}\mathbf{result}\mathbf{will}\mathbf{be}=\frac{1}{5}\times 265$$$$=53\mathbf{ways}$$$$\mathbf{please}\mathbf{check}$$

Commented by mr W last updated on 22/Sep/20

$$correctsir!$$

Commented by PRITHWISH SEN 2 last updated on 22/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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