Maria has three identical apples and three identical oranges. How many ways are there for her to distribute the fruits among her four friends if she doesn't give Jacky any oranges? (Note: Maria takes none of the fruit herself, and some of her friends might not receive any fruit at all.)

 Sep 13, 2018

I will take a crack at this one:


Will consider this in 2 separate steps:

1 - Distributing 3 identical apples among 4 friends:

(3+ 4 -1) C (4 - 1) =6C3 =20 different ways.

2-Distributing 3 identical oranges among 3 friends:

(3 + 3 -1) C (3 - 1) =5C2 =10 different ways.

So, the total =20 + 10 = 30 different ways.

Note: somebody should check this.

 Sep 13, 2018

This was pretty clever thinking actually.  If you had multiplied instead of added you would have gotten it right.  Well done!

Rom  Sep 13, 2018

Suppose Jacky gets 0 apples.


Then we have 3 apples to distribute among 3 friends, and 3 oranges to distribute among 3 friends.


These are independently distributed so the total number of distributions will be the product of the

numbers of distributions of apples and oranges.


3 apples distributed among 3 friends can be done \(\dbinom{3+3-1}{3-1}=\dbinom{5}{2} = 10\) ways

3 oranges is the same so the total number of valid distributions when Jacky gets 0 apples is \(10\cdot 10 = 100\)


Now suppose Jacky gets 1 apple.  We have 2 apples to distribute among 3 friends, and 3 oranges.  This is


\(\dbinom{2+3-1}{3-1}\dbinom{5}{2} = \dbinom{4}{2}\dbinom{5}{2} = 6\cdot 10=60\)


Continuing in the same way there are \(\dbinom{3}{2}\dbinom{5}{2}=30\) ways to distribute the fruit if Jacky gets 2 apples, and

\(\dbinom{2}{2}\dbinom{5}{2}=10\) ways to distribute the fruit if Jacky gets 3 apples

Summing all these up we get \(100+60+30+10=200 \) different ways to distribute the fruit with Jacky getting no oranges

 Sep 13, 2018

Here is the same answer as Rom's but simplified:



 Sep 14, 2018

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