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how many three-digut numbers can be formed from the numbers 1,2,3,4,5,6,7 and 8 if no repetition is allowed?​

Sagot :

[tex] \large\underline \bold{{SOLUTION}}[/tex]

[tex]\longmapsto\rm{P(n,r)= \frac{n!}{(n-r)!}}[/tex]

[tex]\longmapsto\rm{P(8,3)= \frac{8!}{(8-3)!}}[/tex]

[tex]\longmapsto\rm{P(8,3)= \frac{8!}{5!}}[/tex]

[tex]\longmapsto\rm{P(8,3)= \frac{8×7×6×5×4×3×2×1}{5×4×3×2×1}}[/tex]

[tex]\longmapsto\rm{P(8,3)= \frac{8×7×6×\cancel{5×4×3×2×1}}{\cancel{5×4×3×2×1}}}[/tex]

[tex]\longmapsto\rm{P(8,3)= 8×7×6}[/tex]

[tex]\longmapsto\boxed{\rm{P(8,3)= 336}}[/tex]

[tex] \large\underline{ \bold{ANSWER}}[/tex]

  • 336 ways