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A toy store offers 4 different types of toys and 3 different colours for each type. If a customer wants to buy 2 toys of different types and different colours, how many different combinations of toys can the customer choose from?

Option: 1

109
 


Option: 2

108
 


Option: 3

110

 


Option: 4

105


Answers (1)

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First, we need to choose 2 different types of toys out of the 4 available types.
{ }^4 C_2=\frac{4 !}{2 ! 2 !}=6 \text { ways }
Next, we need to choose 2 different colours out of the 3 available colours for each of the 2 chosen types of toys.

{ }^3 C_2 \times{ }^3 C_2=9 \text { ways. }

Finally, we can combine the chosen types and colours of toys in 2 !=2 ways (i.e. we can choose to play with the first toy or the second toy first).

Therefore, the total number of different combinations of toys that the customer can choose from is:
6 \times 9 \times 2=108

So there are 108 different ways for the customer to choose 2 toys of different types and different colours.

Posted by

Pankaj Sanodiya

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