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Aptitude ➤ Permutations And Combinations ➤ Set 1
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.


i. We define 0! = 1.
ii. 4! = (4 x 3 x 2 x 1) = 24.
iii. 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.


i. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
ii. 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:
  nPr = n(n - 1)(n - 2) ... (n - r + 1) =  n!

i. 6P2 = (6 x 5) = 30.
ii. 7P3 = (7 x 6 x 5) = 210.
iii. 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!
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.

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:
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:
nCr = n! = n(n - 1)(n - 2) ... to r factors .
(r!)(n - r)! r!


i. nCn = 1 and nC0 = 1.
ii. nCr = nC(n-r)


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
Question 11
Q11.  From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Question 12
Q12.  In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Question 13
Q13.  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?
Question 14
Q14.  How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Question 15
Q15.  In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

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