Sagot :
[tex] \large\underline \mathcal{{QUESTION:}}[/tex]
how many 5 digits number can be formed from digits 2,3,4,5,6, and 8?
[tex]\\[/tex]
[tex] \large\underline \mathcal{{SOLUTION:}}[/tex]
» 2,3,4,5,6,8 are 6 objects
[tex]\\[/tex]
- Given that n = 6 and r = 5
[tex]\sf{P(n,r)=\frac{n!}{(n-r)!}}[/tex]
[tex]\sf{P(6,5)=\frac{6!}{(6-5)!}}[/tex]
[tex]\sf{P(6,5)=\frac{6!}{1!}}[/tex]
[tex]\sf{P(6,5)=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{1}}[/tex]
[tex]\sf{P(6,5) = 720}[/tex]
[tex]\\[/tex]
[tex] \large\underline \mathcal{{ANSWER:}}[/tex]
- There are 720 ways
[tex] \footnotesize \begin {aligned} \bold{\textsf{Variation Formula :} }\\ \sf \: V_k(n)=\frac{n!}{(n-k)!}\\ \\ \sf \: n = 6 \\ \sf \: k = 5 \\ \\ \sf \: V_5(6)=\frac{6!}{(6-5)!} = \frac{6}{1} \\ \\ \sf \frac{6 \times 5 \times 4 \times 3 \times 2 \times \cancel1}{ \times \cancel 1} = 720 \\ \\ \boxed{\textsf{720 \: ways}}\end{aligned}[/tex]