Tips And Tricks And Shortcuts For Combination

Question 1 (with repetition)

Jannat walks in a candy shop. She has enough money to buy 6 candies. The store has 1 pink, 1, green, and 1 white candy. How many different selections can Jannat make?

Solution:     ^{r + n – 1}C_r ^{= 6 + 3 – 1}C₆ =⁸C₆

We know that,

^nC_r = \frac{n!}{(n-r)! r! }

⁸C₆= \frac{8!}{(8-6)! 6! } = 28

Question 2 (without repetition)

In how many ways can a swimming coach select 3 swimmers from 8 swimmers?

Solution:    

There are eight swimmers out of which three are to be selected

This means ⁸C₃ = \frac{8!}{(8-3)! 3! } = 56

Question 3 (without repetition)

In a group of 10 people, how many different 5-member committees can be formed if it is mandatory to include two specific individuals in each committee?

Solution:

Since two particular people must be included in the committee, we need to select 3 more members to complete the 5-member committee.

The number of ways to select 3 members out of the remaining 10 – 2 = 8 people can be calculated using combinations.

Number of ways to select 3 out of 8 people = ^8C_3=\frac{8!}{3!\times5!}=56

Question 4 (with repetition)

You have four different types of fruits in a basket: apples, oranges, bananas, and grapes. You want to select three fruits from the basket. How many different combinations of three fruits can you make with repetitions allowed?

Solution:

n this case, n = 4 (number of distinct fruits)

r = 3 (number of fruits to be selected).

Using the formula, the number of combinations is:

^{4+3-1}C_3=^6C_3=\frac{6!}{3!\times3!}=20

Question 5 (without repetition)

You have eight friends, and you want to invite three of them to a party. How many different combinations of three friends can you invite?

Solution:

Here, n = 8 (total number of friends) and r = 3 (number of friends to be invited). Applying the combination formula:

^8C_3=\frac{8!}{3!\times(8-3)!}=\frac{8!}{3!\times5!}=56