Counting principle. The ten different letters A, B, C, D, E, F, G, H, I and J are arranged to form a code using ALL ten letters. How many different arrangements are possible?

1. \(10^{10}\)
2. \(3\,628\,800\)
3. \(362\,880\)
4. \(1024\)

Answer: \(3\,628\,800\)

All ten letters are different and all ten are used, so this is a straight arrangement of 10 distinct objects.

\(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)

Tip: Use \(n!\) when every object is used and all are different. \(10^{10}\) would be right only if letters could repeat.

Counting principle. The ten different letters A to J are arranged using all ten letters. In how many of these arrangements are the letters A, B, C and D all next to one another?

1. \(120\,960\)
2. \(5040\)
3. \(40\,320\)
4. \(24\)

Answer: \(120\,960\)

Treat the block ABCD as ONE object. That leaves the block plus the other 6 letters, so 7 objects to arrange.

Arrange the 7 objects: \(7! = 5040\)

Arrange A, B, C, D inside the block: \(4! = 24\)

\(4! \times 7! = 24 \times 5040\)

Tip: The block method: bundle the items that must stay together, arrange the bundles, then arrange inside the bundle.

Counting principle. The ten different letters A to J are arranged at random using all ten letters. What is the probability that an arrangement starts with A, B, C, D in that exact order?

1. \(\approx 0{,}0333\)
2. \(\approx 0{,}00397\)
3. \(\approx 0{,}000198\)
4. \(\approx 0{,}1\)

Answer: \(\approx 0{,}000198\)

Positions 1 to 4 are fixed as A, B, C, D in that one specific order, so there is only 1 way to fill them.

The remaining 6 letters fill the last 6 positions in any order: \(6! = 720\) ways.

\(P = \dfrac{6!}{10!} = \dfrac{720}{3\,628\,800} = \dfrac{1}{5040}\)

Tip: "In that exact order" fixes the arrangement, so the numerator is 1 x 6!, not 4! x 6!.

Counting principle. Six passengers sit in a row of six seats. In how many different ways can they be seated if two particular passengers must sit next to each other?

1. \(720\)
2. \(120\)
3. \(1440\)
4. \(240\)

Answer: \(240\)

Treat the pair as ONE unit. That gives 5 units to arrange (the pair plus the other 4 passengers).

Arrange the 5 units: \(5! = 120\)

The two passengers can swap inside the pair: \(2! = 2\)

\(5! \times 2 = 120 \times 2\)

Tip: Never forget the internal \(2!\) - the pair can sit in either order.

Counting principle. Six passengers, including Mary, take the six seats in a row at random. What is the probability that Mary sits in one of the two end seats?

1. \(\dfrac{1}{3}\)
2. \(\dfrac{1}{6}\)
3. \(\dfrac{1}{2}\)
4. \(\dfrac{2}{5}\)

Answer: \(\dfrac{1}{3}\)

Count the arrangements with Mary at the left end: fix Mary, then arrange the other 5 in \(5!\) ways.

Same for the right end: another \(5!\) ways.

\(P = \dfrac{5! + 5!}{6!} = \dfrac{240}{720}\)

Tip: Quicker: Mary is equally likely to take any of the 6 seats, and 2 of them are end seats, so \(P = \dfrac{2}{6}\).