Zada has to distribute 15 chocolates among 5 of her children sana, ada, jiya, amir and farhan. she has to make sure that sana gets at least 3 and at most 6 chocolates. in how many ways can this be done?
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According to the question Sana must have at least 3 chocolates, that is, she can have 3 or 4 or 5 or 6 chocolates but not less than 3.
When she gets three, remaining ones can be distributed in 11C3 ways = 165 combinations.
When she gets 4, there are 10C3 ways = 120 combinations.
When she gets 5, there are 9C3 ways = 84 combinations.
When she gets 6, there are 8C3 ways = 56 combinations.
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