matthew is planning dinners for the next 3 nights. there’s are 11 meals to choose from. If no meal is repeated, how many different meal arrangements are possible?

The number of different meal arrangements that are possible for this case when meal will be served 3 nights out of 11 meal choices, with no repetition is 990

How many ways k things out of m different things (m ≥ k) can be chosen if order of the chosen things doesn't matter?

We can use combinations for this case,

Total number of distinguishable things is m.

Out of those m things, k things are to be chosen such that their order doesn't matter.

This can be done in total of

[tex]^mC_k = \dfrac{m!}{k! \times (m-k)!}[/tex] ways.

If the order matters, then each of those choice of k distinct items would be permuted k! times.

So, total number of choices in that case would be:

[tex]^mP_k = k! \times ^mC_k = k! \times \dfrac{m!}{k! \times (m-k)!} = \dfrac{m!}{ (m-k)!}\\\\^mP_k = \dfrac{m!}{ (m-k)!}[/tex]

This is called permutation of k items chosen out of m items (all distinct).

For this case, we're given that:

Matthew planning dinner for next 3 nights (so dinner will be served 3 nights from the considerd place).
There are 11 different choices for each meal.
No meal is going to be repeated on those 3 days.

Here, we've to choose 3 meals out of 11 meals for each of 3 days.

The ordering of those 3 meals matter as [tex]m_1, m_2, m_3[/tex] would be different than [tex]m_2, m_1, m_3[/tex] because we got different meals on first and second day.

Thus, we can use permutations here as:

Number of different meal arrangements for this case = [tex]^{11}P_3 = \dfrac{11!}{ (11-3)!} = 11 \times 10 \times 9 = 990[/tex]

Thus, the number of different meal arrangements that are possible for this case when meal will be served 3 nights out of 11 meal choices, with no repetition is 990

Learn more about combinations and permutations here:

https://brainly.com/question/16107928

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