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Sagot :
✏️PERMUTATIONS
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[tex]\underline{\mathbb{PROBLEM:}}[/tex]
- Find the number of distinguishable permutations of the digits of the number 3,4,8,8,3,8.
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[tex]\underline{\mathbb{ANSWER:}}[/tex]
[tex]\qquad\LARGE\rm» \:\: \green{60\:ways}[/tex]
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[tex]\underline{\mathbb{SOLUTION:}}[/tex]
- This 6-digit number can be arranged in 6! ways and remove the other repeating arrangements such that three 8's is arranged in 3! ways and 2 3's is arranged in 2! ways.
[tex]\begin{aligned} & \bold{\color{lightblue}Formula:} \\ & \boxed{\: \rm P = \frac{n!}{n_1! \ n_2! \ ... \ n_k!} \:}\end{aligned} [/tex]
- [tex]\begin{aligned} \rm P = \frac{6!}{3! \ 2!} \end{aligned} [/tex]
- [tex]\begin{aligned} \rm P = \frac{6 \cdot 5 \cdot 4 \cdot \cancel{3!}}{ \cancel{3!} \cdot 2} \end{aligned} [/tex]
- [tex]\begin{aligned} \rm P = \frac{6 \cdot 5 \cdot \cancel4^{\ 2} }{ \cancel2} \end{aligned} [/tex]
- [tex]\rm P = 6 \cdot 5 \cdot 2[/tex]
- [tex]\rm P = 60[/tex]
[tex]\therefore[/tex] There are 60 ways to arrange the number digits of the number 3,4,8,8,3,8.
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