Selecting $15$ pieces of fruit from a bowl containing apples, bananas, oranges, and pears with conditions

Question

A bowl of fruit contains apples, bananas, oranges, and pears (at least $15$ of each kind).

(a) In how many ways can we choose $15$ pieces?

$C(60,15) = 53194089192720$

(b) In how many ways can we choose $15$ pieces, including at least $5$ oranges?

$C(55,10) = 29248649430$

(c) In how many ways can we choose $15$ pieces so that there are at most $4$ bananas?

• Was wondering if my steps were correct and I can't figure out a way to attempt c

Answer 1

First of all, its not mentioned that there are 60 fruits. Actually, there are at least 60,
(at least 15 of each kind, and there are 4 kinds).

Number of integral solutions of $x_1+x_2+...+x_r=n$ is $C(n+r-1,r-1)$ (why?)

Lets denote your fruits as a,b,o and p.

a- You want 15 fruits, no matter which ones you choose.
So, say that you take a apples, b bananas, o oranges and p pears, such that: a+b+o+p=15.
(a,b,o,p can be zero as well)
So, integral solutions of this equation will be: C(18,3)

b- Now, you want 15 fruits, at least 5 of them being oranges.
So, take a apples, b bananas, o+5 oranges and p pears.
a+b+(o+5)+p=15
a+b+o+p=10
Integral solutions: C(13,3)

c- Here it's easier if you make cases like 0 bananas, 1 banana, 2 and 3 and 4.
You have only 1 way of choosing any number of bananas, all alike.

For 0:
a+o+p=15
C(17,2)

For 1:
a+o+p=14
C(16,2)

For 2:
a+o+p=13
C(15,2)

For 3:
a+o+p=12
C(14,2)

For 4:
a+o+p=11
C(13,2)
Final answer will be sum of all: C(17,2)+C(16,2)+C(15,2)+C(14,2)+C(13,2)

Answer 2

A bowl of fruit contains apples, bananas, oranges, and pears, with at least $15$ of each kind. In how many ways can we choose $15$ pieces of fruit?

That is a mighty large bowl.

The condition that there are at least $15$ of each kind means that there at least $4 \cdot 15 = 60$ pieces of fruit in the bowl, not that there are exactly $60$ pieces of fruit. It also means that we may select as many as $15$ pieces of each type of fruit. Therefore, if we let $x_a$, $x_b$, $x_o$, and $x_p$ denote, respectively, the number of apples, bananas, oranges, and pears that are selected, then $$x_a + x_b + x_o + x_p = 15 \tag{1}$$ is an equation in the nonnegative integers.

A particular solution of equation 1 corresponds to the placement of three addition signs in a row of $15$ ones. For instance, $$1 1 1 1 1 + + 1 1 1 1 1 1 1 + 1 1 1 $$ corresponds to the solution $x_a = 5$, $x_b = 0$, $x_o = 7$, and $x_p = 3$, while $$1 1 1 + 1 1 1 1 + 1 1 1 1 + 1 1 1 1$$ corresponds to the solution $x_1 = 3$, $x_b = 4$, $x_o = 4$, and $x_p = 4$. The number of solutions of equation 1 is the number of ways we can place three addition signs in a row of fifteen ones, which is $$\binom{15 + 3}{3} = \binom{18}{3}$$ since we must choose which three of the eighteen positions required for fifteen ones and three addition signs will be filled with addition signs.

In general, if we wish to find the number of solutions of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the nonnegative integers, we must place $n - 1$ addition signs in a row of $k$ ones, which can be done in $$\binom{k + n - 1}{n - 1}$$ since we must select which $n - 1$ of the $k + n - 1$ positions required for $k$ ones and $n - 1$ addition signs will be filled with addition signs.

A selection of $k$ objects from $n$ types of objects in which there are at least $k$ objects of each type from which to select is a combination with repetition problem.

A bowl of fruit contains apples, bananas, oranges, and pears, with at least $15$ of each kind. In how many ways can we choose $15$ pieces of fruit, including at least $5$ oranges?

We must solve the equation $$x_a + x_b + x_o + x_p = 15$$ subject to the restriction that $x_o \geq 5$.

Let $x_o' = x_0 - 5$. Then $x_o'$ is a nonnegative integer. Substituting $x_o' + 5$ for $x_0$ in equation 1 yields \begin{align*} x_a + x_b + x_o' + 5 + x_p & = 15\\ x_a + x_b + x_o' + x_p & = 10 \tag{2} \end{align*} which is an equation in the nonnegative integers with $$\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3}$$ solutions.

A bowl of fruit contains apples, bananas, oranges, and pears, with at least $15$ of each kind. In how many ways can we choose $15$ pieces of fruit, including at most $5$ bananas?

Notice that the negation of at most $4$ is at least $5$. Therefore, the answer is found by subtracting the answer of the second question from the first.

$$\binom{18}{3} - \binom{13}{3}$$

Answer 3

Ad 1.

You have to pick $a$ apples, $b$ bananas, $c$ oranges and $d$ pears, such that $$a+b+c+d=15$$ Fortunately there are $15$ of each fruit, therefore we can use stars and bars method for $n=15$ and $k=4$: $$X=\binom{15+4-1}{15}=816$$

Ad 2.

We take 5 places for oranges and split the rest of it between fruits (i.e. $a+b+c+d=10$). It can be done in $$Y=\binom{10+4-1}{10}=286$$ ways.

Ad 3.

Take the situation, where there are at least 5 bananas. There are 286 situations like that one (it's analogical for the situation with at least 5 oranges). Situation, where there at most 4 bananas is opposite for this one. Therefore there are $$Z=X-Y=816-286=530$$ situations, where there are at most 4 bananas.