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how many different possible arrangements are there if there are 10 people and only 7 chairs are available​

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AGboiigo

[tex] \large\underline \mathcal{{QUESTION:}}[/tex]

how many different possible arrangements are there if there are 10 people and only 7 chairs are available?

[tex]\\[/tex]

[tex] \large\underline \mathcal{{SOLUTION:}}[/tex]

Using the Linear Permutation Formula:

  • Given that n=10 , r=7

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

[tex]\sf{P(10,7)=\frac{10!}{(10-7)!}}[/tex]

[tex]\sf{P(10,7)=\frac{10!}{3!}}[/tex]

[tex]\sf{P(10,7)=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5\times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}}[/tex]

[tex]\sf{P(10,7)=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5\times 4 \times \cancel{3 \times 2 \times 1}}{ \cancel{3 \times 2 \times 1}}}[/tex]

[tex]\sf{P(10,7)=10 \times 9 \times 8 \times 7 \times 6 \times 5\times 4 }[/tex]

[tex]\sf{P(10,7)=604,800}[/tex]

[tex]\\[/tex]

[tex] \large\underline \mathcal{{ANSWER:}}[/tex]

  • There are 604,800 possible arrangements
Chaser27go

[tex]\footnotesize\begin{aligned}\textsf{Variation Formula:}\\ \sf V_k(n)=\frac{n!}{(n-k)!} \\ \\ \textsf{Solution : } \\ \\ \sf \: n = 10 \\ \sf \: k = 7 \\ \\ \sf \: V_7(10)=\frac{10!}{(10-7)!}=\frac{10!}{3!} \\ \\ \sf \: \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times \cancel{ 3 \times 2 \times 1}}{ \cancel{3 \times 2 \times 1}} = 604,800 \\ \\ \boxed{\textsf{604,800 \: Possible}}\end{aligned}[/tex]

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