Well, you could assign a letter to each piece of luggage like so...
A, B, C, D, E, F, G
What you could then do is set it against a table (a configuration table to be precise) with the same letters, and repeat the process again. If the order of these pieces of luggage also has to be taken into account, you'll end up with more configurations.
My answer and workings are below...
35 arrangements without order taken into consideration, because there are 35 ways in which to select 3 objects from the 7 objects.
210 arrangements (35 x 6) when order is taken into consideration.
*There are 6 ways to configure 3 letters.
Alternative way to solve the problem...
Produce Pascal's triangle. If you want to know how many ways in which you can choose 3 objects from 7, select (7 3) in Pascal's triangle which is equal to 35. Now, there are 6 ways in which to configure 3 objects if you are concerned about order.
BC is congruent to DC or BA is congruent to DA, depending on what point marks the intersection point of the two lines.
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Answer:
Step-by-step explanation:
<u>This is going to be an arithmetic progression:</u>
- 48500, 48500 + 1200, 48500 + 1200*2, ...
<u>So the first term is </u>
<u>and the common difference is </u>
<u>Sum of AP is:</u>
- S = (a₁ + aₙ)*n/2 = (2a₁ + (n - 1)d)*n/2
<u>Substitute and solve for n:</u>
- (2*48500 + (n - 1)*1200)*n/2 = 321000
- (97000 + 1200n - 1200)n = 642000
- 1200n² + 95800n - 642000 = 0
- 6n² + 479n - 3210 = 0
<u>Solving we get a positive root of:</u>
- n ≈ 6.21 and it rounds up to 7 years