This question is a good example where both
OR and
AND events are used. Let's see how such events are treated when finding the number of ways of an event happening.
Given InfoWe are given that a restaurant in its dessert platter offers 6 kinds of cheese and 2 kinds of fruits. We are asked to find the number of ways a dessert platter can be offered if the platter contains equal kinds of fruits and cheese.
ApproachWe observe here that there are lesser kinds of fruits than the cheese. Hence the number of ways the platter can be offered is limited by the number of fruits.
Since a platter would be incomplete without any of the fruit or cheese, for the event to occur( i.e. platter to be served) a platter should have fruit
AND cheese.
Also, there can be a case where a platter may have 1 fruit
AND 1 cheese
OR 2 fruits
AND 2 cheese. Hence, the event(i.e. platter to be served) can occur in either of the ways.
Number of ways the platter can be served = Number of ways of picking up 1 fruit out of 2 fruits
AND number of ways of picking up 1 cheese out of 6 cheese
OR Number of ways of picking up 2 fruits out of 2 fruits
AND number of ways of picking up 2 cheese out of 6 cheese
Working OutNumber of ways in which 1 fruit can be picked out of 2 fruits = \(2_{c_{1}}\)
Number of ways in which 1 cheese can be picked out of 6 cheese = \(6_{c_{1}}\)
Hence number of ways in which a platter having 1 fruit and 1 cheese can be served = \(2_{c_{1}}\) \(*\) \(6_{c_{1}}\) \(= 2 * 6 = 12\)
Similarly, Number of ways in which 2 fruits can be picked out of 2 fruits = \(2_{c_{2}}\)
Number of ways in which 2 cheese can be picked out of 6 cheese = \(6_{c_{2}}\)
Hence number of ways in which a platter having 2 fruits and 2 cheese can be served = \(2_{c_{2}}\) \(*\) \(6_{c_{2}}\) \(= 1 * 15 = 15\)
Therefore total number of ways of serving a dessert platter \(= 12 + 15 = 27\)
Hope this helps
Regards
Harsh
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63% (01:27) correct 
