## Fundamental Principle of Counting

**Multiplication**: If there are two operations can be performed in m, n ways respectively and both operations has to be performed mutually then possible operations can be performed in m×n ways. Suppose a die is cast and after that a coin is tossed then number of possible outcomes will be 6×2=12.**Addition**: If there are two operations can be performed in m, n ways respectively and both operations are independent then possible operations can be performed in m+n ways. Suppose it is asked whether to cast a die or toss a coin then number of possible outcomes will be 6+2=8.

**Q1: There are 6 trains between Bangalore and Mumbai, how many ways a person can go and come back by a different train?**

**Answer**

Number of ways = Ways person can go (6 trains) × Ways person can come (5 trains since cannot come by same train)

Number of ways = 6×5 = 30

**Q2: In how many ways a 2-digit number can be formed from given digits 4, 6, 8, 9 and digits should not repeat?**

**Answer**

Number of ways = 4 (choose any digit from given set) × 3 (choose any digit from given set except one already chosen)

Number of ways = 4×3 = 12

## Permutations

Each different “arrangement” out of given things by taking some or all at a time is called “permutation”.

**Q3: How many ways A, B and C can be arranged in group of 2?**

**Answer**

A, B and C can be arranged as AB, BA, BC, CB, AC, CA so 6 ways

Or ^{3}P_{2} = 3! ÷ (3-2)! = (3×2×1) ÷ (1) = 6

**Q4: Find the number of ways which 12 books can be arranged on a shelf that two particular books shall not be together?**

**Answer**

Number of ways 12 books can be arranged = ^{12}P_{12} = 12!

Number of ways 12 books can be arranged when two particular books are always together

= Number of ways 11 books can be arranged (consider 2 books together as one) × Number of ways 2 books can be arranged (those two particular books can be arranged in !2 ways)

= ^{11}P_{11} × ^{2}P_{2}

= 11! × 2!

Now Number of ways 12 books can be arranged when two particular books are never together

= 12! - 11! × 2!

= 11! × (12 - 2)

= 399168000

### Circular Permutations

Since things will be arranged in form of circle so there will be no start or end so number of ways n things can be arranged in a circle are (n-1)!

**Q5: Find the number of ways 5 men and 2 women can be arranged on a round table where women shall not sit together?**

**Answer**

Number of ways 5 men can be arranged in circular table = 4!

Number of ways 2 women can be arranged in between 5 men = ^{5}P_{2}

Total permutations = 4! × ((5!) ÷ (5-2)!) = 480

## Combinations

Each different “combination” out of given things by taking some or all at a time is called “combination”.

**Q6: A committee of 5 persons to be formed from 6 men and 4 women, in how many ways it can be done when at most two women are included?**

**Answer**

Number of ways = (2 women AND 3 men) OR (3 women AND 2 men) OR (4 women AND 1 men)

= (^{4}C_{2} × ^{6}C_{3}) + (^{4}C_{3} × ^{6}C_{2}) + (^{4}C_{4} × ^{6}C_{1})

= 186

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