Vtyjkyuk

For her birthday, Clara wanted a balloon bouquet. If there are four colors to choose from and her bouquet will have four balloons, how many combinations of colored balloons are possible?

Imagine that an urn contains four pieces of paper, and each piece of paper has the name of a color written on it. We select four items from this urn, with replacement, and order does not matter. Each time we select a piece of paper, we add a balloon of the corresponding color to our bouquet.

The "stars and bars" formula tells us that the number of possible outcomes is $binom73$.

Alternative viewpoint: let $x_i$ be the number of balloons of color $i$. The numbers$x_i$ satisfy
$$
x_1+x_2+x_3+x_4=4.
$$
The number of solutions to this equation, where each $x_i$ is a nonnegative integer, is $binom73$, as can be shown using a "stars and bars" argument.

Further details:

Here is the "stars and bars" approach to counting the number of possible outcomes. We are making $4$ selections from the urn, so let's write down $4$ stars:
$$
star , star , star , star
$$
Next, we need to decide how many of the selections will be for color 1, how many for color 2, how many for color 3, and how many for color 4. We can make this decision by drawing $3$ vertical bars, like this:
$$
star , mid , star , star , mid , mid , star
$$
The placement of vertical bars in this example means that the first color was selected $1$ time, the second color was selected $2$ times, the third color was selected $0$ times, and the fourth color was selected $1$ time. Other placements of bars correspond to other possible outcomes, for example:
$$
star , mid , star , star , star , mid , mid
$$
In this outcome, the first color has been selected $1$ time, the second color has been selected $3$ times, and the third and fourth colors have been selected $0$ times.

Notice that each possible outcome corresponds to a sequence of 7 symbols, where each symbol is either a star or a bar. Thus, to specify a particular outcome, we only need to choose which $3$ of the $7$ symbols will be bars. (The remaining symbols will be stars.) This shows that the number of possible outcomes is $binom73$.

Hint:

Discern cases: $abcd, abcc, aabb, aaab, aaaa$

Here e.g. $abcc$ stands for the situation of a bouquet where exactly $3$ colors are represented.

Further e.g. $aaab$ stands for the situation of a bouquet where one color is represented exactly $3$ times.

For every case find out how many possibilities are there.

Especially in case $aabb$ watch out for double counting.