If an operation can be performed in @@m@@ different ways and following which a second operation can be performed in @@n@@ different ways, then the two operations in succession can be performed in @@m × n@@ different ways.

**2. Addition Theorem (Fundamental Principles of Counting)**

If an operation can be performed in @@m@@ different ways and a second independent operation can be performed in @@n@@ different ways, either of the two operations can be performed in @@(m+n)@@ ways.

**3. Factorial**

Let @@n@@ be a positive integer. Then @@n@@ factorial can be defined as

@@n!=n(n-1)(n-2)\cdots 1@@

**Examples**

@@5!=5×4×3×2×1=120\\~\\3!=3×2×1=6@@

**Special Cases**

@@0! = 1\\~\\1! = 1@@

Permutations are the different arrangements of a given number of things by taking some or all at a time.

**Examples**

All permutations (or arrangements) that can be formed with the letters a, b, c by taking three at a time are (abc, acb, bac, bca, cab, cba)

All permutations (or arrangements) that can be formed with the letters a, b, c by taking two at a time are (ab, ac, ba, bc, ca, cb)

Each of the different groups or selections 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 girls P, Q, R. Then, possible combinations are PQ, QR and RP. *(Note that PQ and QP represent the same selection.)*

Suppose we want to select three out of three girls P, Q, R. Then, only possible combination is PQR

**6. Difference between Permutations and Combinations and How to identify them**

Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination.

Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not.

If order is important, PQ will be different from QP, PR will be different from RP and QR will be different from RQ

If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ

Hence,

If the order is important, problem will be related to permutations.

If the order is not important, problem will be related to combinations.

For permutations, the problems can be like "What is the number of permutations the can be made", "What is the number of arrangements that can be made", "What are the different number of ways in which something can be arranged", etc.

For combinations, the problems can be like "What is the number of combinations the can be made", "What is the number of selections the can be made", "What are the different number of ways in which something can be selected", etc.

Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures, distribution of items (there are exceptions for this) etc will be related to combinations.

The term repetition is very important in permutations and combinations. Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R.

If repetition is allowed, the same object can be taken more than once to make a sample. i.e., PP, QQ, RR can also be considered as possible samples.

If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples.

Normally repetition is not allowed unless mentioned specifically.

**8. Number of permutations of n distinct things taking r at a time **

Number of permutations of n distinct things taking r at a time can be given by^{n}P_{r} = @@\dfrac{n!}{(n - r)!}@@ @@= n(n-1)(n-2)...(n-r+1) \quad @@ where @@0 \le r \le n@@

**Special Cases**^{n}P_{0} = 1^{n}P_{r} = 0 for @@r \gt n@@

^{n}P_{r} is also denoted by P(n,r). ^{n}P_{r} has importance outside combinatorics as well where it is known as the falling factorial and denoted by (n)_{r} or n^{r}

**Examples**

^{8}P_{2} = 8 × 7 = 56^{5}P_{4}= 5 × 4 × 3 × 2 = 120

**9. Number of permutations of n distinct things taking all at a time **

Number of permutations of n distinct things taking them all at a time

= ^{n}P_{n} = n!

**10. Number of Combinations of n distinct things taking r at a time**

Number of combinations of n distinct things taking r at a time ( ^{n}C_{r}) can be given by ^{n}C_{r} = @@\dfrac{n!}{(r!)(n - r)!}@@ @@=\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}\quad @@ where @@0 \le r \le n@@

**Special Cases**

^{n}C_{0} = 1^{n}C_{r} = 0 for @@r \gt n@@

^{n}C_{r} is also denoted by C(n,r). ^{n}C_{r} occurs in many other mathematical contexts as well where it is known as binomial coefficient and denoted by @@\dbinom{n}{r}@@

**Examples**

^{8}C_{2} = @@\dfrac{8 \times 7}{2 \times 1}@@ = 28

^{5}C_{4}= @@\dfrac{5 \times 4 \times 3 \times 2}{4 \times 3 \times 2 \times 1}@@ = 5

1st ring can go into any of the 4 fingers

2nd ring can go into any of the 4 fingers

3rd ring can go into any of the 4 fingers

4th ring can go into any of the 4 fingers

5th ring can go into any of the 4 fingers

6th ring can go into any of the 4 fingers

therefore answer is 4*4*4*4*4*4 = 4^6

Suppose, A,B,C,D,E,F are the rings. Then, these 360 arrangements are the arrangements like ABCD, ABDC, ABCE etc.

But, this is not the required count as per the question. Answer is 4^6=4096. There 4096 arrangements also include the arrangements having more than one ring in the same finger and/or no rings in some fingers.

e.g. I want to pick up 4 number (here 4 number is dynamic) n1n2n3n4

and again for each number position i.e. 1 to 6 possible option will be dynamic.

1 st number could be = 1,2,3

2nd number could be = 1

3rd number could be = 1,2

4th number could be = 5,6,7

any algorithm get all those permutation?

I have a question, if we have 6 white balls, 3 blue balls and 2 Red balls and we have to pick a selection 4?

a) how many combinations it will be?

b) if always 2 white have to be selected from 4 then how many combinations possible?

So, 6 + 3 + 2 = 11.

11C4 = $\dfrac{11!}{4!×(11-4)!}=11×10×3=330$

b) The 2nd part i think you didn't give a clarity. Out of 4 if at least two must be white.

then,

picking 4 whites out of 6 whites = 6C4

picking 3 whites with 1 blue or red = 6C3 × 3C1 × 2C1

picking 2 whites with 2 blue = 6C2 × 3C2

picking 2 whites with 2 red = 6C2 × 2C2

picking 2 whites with 1 red 1 blue = 6C2 × 3C1 × 2C1

Now multiply all.

1) we can select any 4 balls out of (6B,3B,2R) so total combinations are 11C4

2)now in the second part exactly 2 balls are white so combinations are 6C2*5C2

Post Your Comment