 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.
 4! = (4 x 3 x 2 x 1) = 24.
 5! = (5 x 4 x 3 x 2 x 1) = 120.
 Permutations:The different arrangements of a given number of things by taking some or all at a time, are called permutations.Examples:
 All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba,ac, ca, bc, cb).
 All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)
 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}P_{2} = (6 x 5) = 30.
 ^{7}P_{3} = (7 x 6 x 5) = 210.
 Cor. number of all permutations of n things, taken all at a time = n!.
 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}!))  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:
 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.
 All the combinations formed by a, b, c taking ab, bc, ca.
 The only combination that can be formed of three letters a, b, c taken all at a time is abc.
 Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
 Note that ab ba are two different permutations but they represent the same combination.
 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.
2.  In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?  
Answer: Option C Explanation: 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 how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?  

4.  Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?  
Answer: Option C Explanation: Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters.
Required number of ways = (210 x 120) = 25200. 
5.  In how many ways can the letters of the word ‘LEADER’ be arranged?  
Answer: Option C Explanation: The word ‘LEADER’ contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Answer: Option C Explanation:

9.  A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?  
Answer: Option C Explanation: We may have(1 black and 2 nonblack) or (2 black and 1 nonblack) or (3 black).

10.  In how many different ways can the letters of the word ‘DETAIL’ be arranged in such a way that the vowels occupy only the odd positions?  
Answer: Option C Explanation: There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5. Number of ways of arranging the vowels = ^{3}P_{3} = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = ^{3}P_{3} = 3! = 6. Total number of ways = (6 x 6) = 36. 
11.  In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?  
Answer: Option A Explanation:

12.  How many 4letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?  
Answer: Option C Explanation: ‘LOGARITHMS’ contains 10 different letters.

13.  In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?  
Answer: Option C Explanation: In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Required number of words = (10080 x 12) = 120960. 
14.  In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?  
Answer: Option B Explanation: The word ‘OPTICAL’ contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5! = 120 ways. The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. 