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