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In the given problem, apply your knowledge in algebraic expression and basic mathematics especially in permutation and combination.

Situation

On her way home from work Jane's mother take a jeepney ride. Following the safety protocol of one seat apart,

Question

a. In how many ways will be she seated together with other 7 passengers?

Let's say _ is the vacant seat

_, _, _, _, _, _, _

7, 6, 5, 4, 3, 2, 1

→ 7 × 6 × 5 × 4 × 3 × 2 × 1 = 7!

7! = 5 040, ans.

b. Would it be possible that she and other two passengers only at a time be arranged? how many sitting arrangements would there be?

Only at a time means, they seat on one part only.

_, _, _, _, _, _, _

Let say x, y, z are those three passengers including Jane's mother.

If they seated at x, y, z, _, _, _, _, → x, y, z, 4, 3, 2, 1 and x y z can be arranged at y z x and z y x and etc.

Thus, 3, 2, 1 for x, y, z passengers and remaining of 4, 3, 2, 1 for other passengers.

3 × 2 × 1 × 4 × 3 × 2 × 1 = 144

It is also equal to nPr = 3P3 for x, y, z passengers, and nPr = 4P4

nPr = 3P3 × 4P4

nPr = 144, ans.

c. Can you draw at least one possible arrangement of Jane's mother and other two passengers ?

Let say x is Jane's mother and y and z for two other passengers. 1, 2, 3, 4 means the four passengers who occupy the remaining four seats.

x, y, z, 1, 2, 3, 4 ans.

Learn more about permutation and combination here at brainly.ph/question/13262774

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