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When order matters and an object can be chosen more than once then the number of
permutations is:
where $n$ is the number of objects from which you can choose and $r$ is the number to be chosen.
For example, if you have the letters A, B, C, and D and you wish to discover the number of ways of arranging them in three letter patterns (trigrams) you find that there are ${4}^{3}$ or 64 ways. This is because for the first slot you can choose any of the four values, for the second slot you can choose any of the four, and for the final slot you can choose any of the four letters. Multiplying them together gives the total.
In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Its simplest version reads
Whenever $n$ is a positive integer, the numbers
are the binomial coefficients (the coefficients in front of the powers).
For example, here are the cases n = 2, n = 3 and n = 4:
The coefficients form a triangle, where each number is the sum of the two numbers above it:
This formula, and the triangular arrangement of the binomial coefficients, are often attributed to Blaise Pascal who described them in the 17th century. It was, however, known to the Chinese mathematician Yang Hui in the 13th century, the earlier Persian mathematician Omar KhayyÃ¡m in the 11th century, and the even earlier Indian mathematician Pingala in the 3rd century BC.
The number plate on a car consists of any 3 letters of the alphabet (excluding the vowels and 'Q'), followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that the number plate starts with a 'Y' and ends with an odd digit?
The number plate starts with a 'Y', so there is only 1 choice for the first letter, and ends with an even digit, so there are 5 choices for the last digit (1, 3, 5, 7, 9).
Use the counting principle. For each of the other letters, there are 20 possible choices (26 in the alphabet, minus 5 vowels and 'Q') and 10 possible choices for each of the other digits.
Number of events = $1\times 20\times 20\times 10\times 10\times 5=200\phantom{\rule{0.277778em}{0ex}}000$
Use the counting principle. This time, the first letter and last digit can be anything.
Total number of choices = $20\times 20\times 20\times 10\times 10\times 10=8\phantom{\rule{0.277778em}{0ex}}000\phantom{\rule{0.277778em}{0ex}}000$
The probability is the number of events we are counting, divided by the total number of choices.
Probability = $\frac{200\phantom{\rule{0.277778em}{0ex}}000}{8\phantom{\rule{0.277778em}{0ex}}000\phantom{\rule{0.277778em}{0ex}}000}=\frac{1}{40}=0,025$
Show that
Method 1: Expand the factorial notation.
Cancelling the common factor of $(n-1)\times (n-2)\times ...\times 2\times 1$ on the top and bottom leaves $n$ .
So $\frac{n!}{(n-1)!}=n$
Method 2: We know that $P(n,r)=\frac{n!}{(n-r)!}$ is the number of permutations of $r$ objects, taken from a pool of $n$ objects. In this case, $r=1$ . To choose 1 object from $n$ objects, there are $n$ choices.
So $\frac{n!}{(n-1)!}=n$
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