If n distinct items are arranged, the number of ways they can be arranged so that they do not occupy their intended spot is $$D = n!$$($$ \dfrac{1}{0!}$$ - $$\dfrac{1}{1!}$$ + $$\dfrac{1}{2!}$$ - $$\dfrac{1}{3!}$$ + .... + $$\dfrac{(-1)^{n}}{n!}$$)
For, example, Derangements of 4 will be D(4) = $$4!\left(1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}\right)=24\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}\right)=24\left(\dfrac{12-4+1}{24}\right)=9$$
D(1) = 0, D(2) = 1, D(3) = 2, D(4) = 9, D(5) = 44, and D(6) = 265
Both of Permutations & Combinations and Probability are extremely important topics in CAT. We have combined both of them together because improving in P&C would help you get better in Probability. After number systems, this is the second most important topic for CAT preparation. Many students find learning P&C and probability a nightmarish experience. Many even skip this topic in frustration. We think under-preparing in P&C and probability is one of the biggest mistakes you could make in CAT prep. This topic can be the most rewarding topic in quant section. Unlike number systems questions, these questions generally take lesser time to solve. Also, they are generally fairly basic in nature. And the best part is- the more questions you solve, the better you will get at this subject. So look through the formula list a few times and understand the formulae. But the best way to tackle this subject is by solving questions. Solve as many as you can- solve so many from this topic that you start to see that all of them are generally variations of the same few themes that are listed in the formula list.
To get the rank of a word in the alphabetical list of all permutations of the word, start with alphabetically arranging the n letters. If there are x letters higher than the first letter of the word, then there are at least x*(n-1)! Words above our word. After removing the first affixed letter from the set if there are y letters above the second letter then there are y*(n-2)! words more before your word and so on. So rank of word = x*(n-1)! + y*(n-2)! + .. +1
If x items out of n items are repeated, then the number of ways of arranging these n items is $$\dfrac{n!}{x!}$$ ways.
If x items, y items and z items are repeated within n items, they can be arranged in $$\dfrac{n!}{x!y!z!}$$ ways.
Let $$E_{1}, E_{2}, E_{3}...$$ be mutually exclusive and collectively exhaustive events each with a probability $$p_{1}, p_{2}, p_{3}...$$ of occurring. Let B be another event of non-zero probability such that probability of B given $$E_{1}$$ is $$q_{1}$$, B given $$E_{2}$$ is $$q_{2}$$ etc. By Bayes theorem: $$$P(E_{i}/B) = \frac{p_{i}q_{i}}{\sum_{j=1}^{n}p_{j}q_{j}}$$$