1. Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

Number of all permutations of n things, taken r at a time, is given by:

The number of all combinations of n things, taken r at a time is:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

2. 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!

3. Number of Combinations:The number of all combinations of n things, taken r at a time is:

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

1.Factorial Notation:

We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

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)

All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)

6P2 = (6 x 5) = 30.

7P3 = (7 x 6 x 5) = 210

number of all permutations of n things, taken all at a time = n!

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!)

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.

Note:

nCn = 1 and nC0 = 1.

nCr = nC(n - r)

1. 11C4 = (11*10*9*8)/(1*2*3*4) = 330

2. 16C13 = 16C(16-13) = 16C3 =(16*15*14)/(1*2*3) = 560

- 1. Problems on Train
- 2. Time and Work
- 3. Percentage
- 4. Time and Distance
- 5. Problems on ages
- 6. Profit and Loss
- 7. Average
- 8. Permutation and Combination
- 9. Problems on H.C.F and L.C.M
- 10. Square Root and Cube Root
- 11. Chain Rule
- 12. Alligation or Mixture
- 13. Stocks and Shares
- 14. Banker's Discount
- 15. Simple Interest
- 16. Compound Interest
- 17. Partnership
- 18. Height and Distance
- 19.Calendar

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