# Exam Mission

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1. Factorial Notation:
Let n be a positive integer. Then, factorial n, denoted n! is defined as:
n! = n(n - 1)(n - 2) ... 3.2.1.
Examples:
1. We define 0! = 1.
2. 4! = (4 x 3 x 2 x 1) = 24.
3. 5! = (5 x 4 x 3 x 2 x 1) = 120.
2. Permutations:
The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Examples:
1. All permutations (or arrangements) made with the letters abc by taking two at a time are (abbaaccabccb).
2. All permutations made with the letters abc taking all at a time are:
( abcacbbacbcacabcba)
3. Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
 nPr = n(n - 1)(n - 2) ... (n - r + 1) = n! (n - r)!
Examples:
1. 6P2 = (6 x 5) = 30.
2. 7P3 = (7 x 6 x 5) = 210.
3. Cor. number of all permutations of n things, taken all at a time = n!.
4. An Important Result:
If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.
 Then, number of permutations of these n objects is = n! (p1!).(p2)!.....(pr!)
5. Combinations:
Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Examples:
1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note: AB and BA represent the same selection.
2. All the combinations formed by abc taking abbcca.
3. The only combination that can be formed of three letters abc taken all at a time is abc.
4. Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
5. Note that ab ba are two different permutations but they represent the same combination.
6. Number of Combinations:
The number of all combinations of n things, taken r at a time is:
 nCr = n! = n(n - 1)(n - 2) ... to r factors . (r!)(n - r)! r!
Note:
1. nCn = 1 and nC0 = 1.
2. nCr = nC(n - r)
Examples:
 i.   11C4 = (11 x 10 x 9 x 8) = 330. (4 x 3 x 2 x 1)
 ii.   16C13 = 16C(16 - 13) = 16C3 = 16 x 15 x 14 = 16 x 15 x 14 = 560. 3! 3 x 2 x 1