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:
We define 0! = 1.
3! = (3 x 2 x 1) = 6.
2. Permutations:
The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Examples:
All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)
3. Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
^{n}P_{r} = n(n  1)(n  2) ... (n  r + 1) = n!/(n  r)!
Examples:
^{6}6P_{2} = (6 x 5) = 30.
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 p_{1} are alike of one kind; p_{2} are alike of another kind; p_{3} are alike of third kind and so on and p_{r} are alike of r^{th} kind,
such that (p_{1} + p_{2} + ... p_{r}) = n.
Then, number of permutations of these n objects is = n!/(p_{1}!).(p_{2}!).....(p_{r}!)
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:
i. 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.
ii. All the combinations formed by a, b, c taking ab, bc, ca.
iii. The only combination that can be formed of three letters a, b, c taken all at a time is abc.
iv. Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
v. 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:
^{n}C_{r} = n!/(r!)(n  r)! = n(n  1)(n  2) ... to r factors/r! .
Note:
^{n}C_{n} = 1 and ^{n}C_{0} = 1.
^{n}C_{r} = ^{n}C_{(n  r)}
Examples:
i. ^{11}C_{4} = (11 x 10 x 9 x 8)/(4 x 3 x 2 x 1) = 330.
ii. ^{16}C_{13} = ^{16}C_{(16  13)} = ^{16}C_{3} = 16 x 15 x 14/3! = 16 x 15 x 14/3 x 2 x 1 = 560.
1. 

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (^{7}C_{3} x ^{6}C_{2}) + (^{7}C_{4} x ^{6}C_{1}) + (^{7}C_{5}).
Required number of ways = [(7 x 6 x 5 )/(3 x 2 x 1)] x [(6 x 5)/(2 x 1)] + (^{7}C_{3} x ^{6}C_{1}) + (^{7}C_{2}).
Required number of ways = 525 + [(7 x 6 x 5)/(3 x 2 x 1)] x 6 + [(7 x 6)/(2 x 1)].
Required number of ways = 525 + 210 + 21 = 756
2. 

The word LEADING has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
3. 

In the word CORPORATION, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.
4. 

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (^{7}C_{3} x ^{4}C_{2}) = [(7 x 6 x 5)/(3 x 2 x 1)] x [(4 x 3)/(2 x 1) = 210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5!
Number of ways of arranging 5 letters among themselves = 5 x 4 x 3 x 2 x 1
Number of ways of arranging 5 letters among themselves = 120.
Required number of ways = (210 x 120) = 25200.
5. 

The word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways = 6!/((1!)(2!)(1!)(1!)(1!)) = 360.
6. 

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (^{6}C_{1} x ^{4}C_{3}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{4})
Required number of ways = (^{6}C_{1} x ^{4}C_{1}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{2})
Required number of ways = (6 x 4) + [(6 x 5)/(2 x 1)] x [(4 x 3)/(2 x 1)] + [(6 x 5 x 4)/(3 x 2 x 1)] x [4] + [(6 x 5)/(2 x 1)]
Required number of ways = (24 + 90 + 80 + 15)
Required number of ways = 209
7. 

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is one way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
8. 

The word BIHAR contains 5 different letters.
Required number of words = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120
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