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Question-205083




Question Number 205083 by BaliramKumar last updated on 07/Mar/24
Answered by A5T last updated on 07/Mar/24
1×1×30;1×2×15;1×3×10;1×5×6;  2×3×5; There are 5 ways upto permutation  but if order matters there are, 4×3!+((3!)/(2!))=27
$$\mathrm{1}×\mathrm{1}×\mathrm{30};\mathrm{1}×\mathrm{2}×\mathrm{15};\mathrm{1}×\mathrm{3}×\mathrm{10};\mathrm{1}×\mathrm{5}×\mathrm{6}; \\ $$$$\mathrm{2}×\mathrm{3}×\mathrm{5};\:{There}\:{are}\:\mathrm{5}\:{ways}\:{upto}\:{permutation} \\ $$$${but}\:{if}\:{order}\:{matters}\:{there}\:{are},\:\mathrm{4}×\mathrm{3}!+\frac{\mathrm{3}!}{\mathrm{2}!}=\mathrm{27} \\ $$
Commented by BaliramKumar last updated on 07/Mar/24
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$$$ \\ $$

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