Permutation and Combination: Solved Examples to enhance better understanding


1.  In how many ways can 7 men and 7 women sit down at a round table in such a way that no 2 men sit next to each other?

Solution: Clearly, the men and women are to sit in alternate positions. When we fix the position of one man, say at the top, the rest 6 men can be arranged in 6! ways.

The 7 women can occupy 7 position in 7! ways. Therefore the total number of ways = 7! * 6!

2. How many eight letter words can be formed from the letters of the word “Courtesy”, beginning with C and ending with Y?

Solution: The first and the last place are fixed with C and Y respectively. The remaining 6 letters can be arranged in 6! Ways. Therefore, the total number of ways = 720.

3. Out of 8 gentlemen and 5 ladies, a committee of 5 is to be formed. Find the number of ways in which this can be done so as to include at least 3 ladies.

Solution: The possible combinations fulfilling the given conditions are:

3 ladies and 2 gentlemen, 4 ladies and 1 gentlemen, and all the 5 ladies.

a)      3 ladies and 2 gentlemen can be selected in: ⁵C₃ X ⁸C₂ = 280 ways

b)      4 ladies and 1 gentleman can be selected in: ⁵C₄ X ⁸C₁ = 40 ways

c)       5 ladies can be selected in 1 ways

The total number of ways the committee can be formed = 321 ways

4. How many different 7 digit telephone numbers can be formed from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9?

Solution: Since 0 cannot be the first digit in the 7-digit telephone number, there are 9 different possibilities for the first position.

The remaining 6 digits can be selected from the 10 digits in 10⁶ ways. The total number of ways = 10⁶x 9

(Remember that in telephone numbers, digits can be repeated. Hence, for every single digit, except for the first position, there are 10 different possibilities)

5. From 3 mangoes, 4 apples and 2 oranges, how many selections of fruits can be made, taking at least one of each kind?

Solution: You may be tempted to adopt the approach we took in the previous question. However, there is no condition given regarding the maximum number of fruits in selection. Hence, we have to try a different technique:

Number of ways in which mangoes can be selected = 2³. But this also includes the case where all three mangoes are not selected. Hence, number of ways in which at least one mangoes is selected = (2³- 1) = 7

Similarly, number of apples can be selected in (2⁴ – 1) ways = 15
number of oranges can be selected in (2² – 1) ways = 3

The total number of ways is 15 x 3 x 7 = 315 ways (we have to multiply and not simply add)


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