Distinct Terms in Multinomial Expansion

This problem asks for the total count of distinct terms that appear when the expression $(p + q + r + s)^n$ is fully expanded. Here, $p, q, r, s$ are variables, and $n$ is a positive integer ($n \in \mathbb{N}$).

Multinomial Expansion Basics

When we expand an expression like $(p + q + r + s)^n$, each term is formed by selecting variables from the $n$ available factors. Each resulting term will generally be of the form $C \cdot p^{a_1} q^{a_2} r^{a_3} s^{a_4}$, where $C$ is a numerical coefficient.

The key conditions related to the exponents in any such term are:

• The exponents $a_1, a_2, a_3, a_4$ must be non-negative integers (meaning they can be $0, 1, 2, \dots$).
• The sum of these exponents must equal the total power $n$:

  $a_1 + a_2 + a_3 + a_4 = n$

The number of distinct terms in the expansion is precisely the number of different combinations of non-negative integer exponents $(a_1, a_2, a_3, a_4)$ that satisfy this sum condition.

Stars and Bars Method for Counting Terms

Finding the number of non-negative integer solutions to an equation like $a_1 + a_2 + a_3 + a_4 = n$ is a standard problem in combinatorics. The "stars and bars" technique provides a direct way to find this count.

We can think of this problem as distributing $n$ identical items (the 'stars', representing the total power) into $k=4$ distinct bins (representing the variables $p, q, r, s$). To divide the items into $k$ bins, we need $k-1$ separators (the 'bars').

In our case, we have $n$ stars and $k-1 = 4-1 = 3$ bars. For example, an arrangement like `***|||` would correspond to $a_1=3, a_2=0, a_3=0, a_4=0$ (if $n=3$). An arrangement like `*|**||` would correspond to $a_1=1, a_2=2, a_3=0, a_4=0$ (if $n=3$).

The total number of possible arrangements of these $n$ stars and $k-1$ bars is equivalent to choosing the positions for the $k-1$ bars (or the $n$ stars) from a total of $n + (k-1)$ available positions.

The general formula derived from the stars and bars method for non-negative integer solutions is:

$ \binom{n + k - 1}{k - 1} $

Alternatively, this can be written as:

$ \binom{n + k - 1}{n} $

Calculation of Distinct Terms

For the given expansion $(p + q + r + s)^n$, we have $k=4$ variables. Substituting $k=4$ into the stars and bars formula:

Number of distinct terms = $ \binom{n + 4 - 1}{4 - 1} $

This simplifies to:

$ \binom{n + 3}{3} $

Now, we need to calculate the value of this binomial coefficient:

$ \binom{n + 3}{3} = \frac{(n+3)!}{3!(n+3-3)!} $

$ = \frac{(n+3)!}{3!n!} $

We can expand the factorial $(n+3)!$ in the numerator as $(n+3) \times (n+2) \times (n+1) \times n!$. Also, $3! = 3 \times 2 \times 1 = 6$.

$ = \frac{(n+3) \times (n+2) \times (n+1) \times n!}{ 6 \times n!} $

By cancelling the $n!$ term from the numerator and the denominator, we get:

$ = \frac{(n+1)(n+2)(n+3)}{6} $

Verification with Options

The result we obtained from our calculation is $\frac{(n+1)(n+2)(n+3)}{6}$. Let's examine the provided options:

• Option 1: $n+1$
• Option 2: $\frac{(n+1)(n+2)}{2}$
• Option 3: $\frac{(n+1)(n+2)(n+3)(n+4)}{24}$
• Option 4: $\frac{(n+1)(n+2)(n+3)}{6}$

Comparing our derived formula with these options, we find that our result matches exactly with Option 4.

Conclusion on Term Count

The number of distinct terms in the expansion of $(p + q + r + s)^n$ depends on the number of ways we can choose non-negative integer exponents for $p, q, r,$ and $s$ such that they sum up to $n$. Applying the stars and bars combinatorial method, we have determined this count to be $\frac{(n+1)(n+2)(n+3)}{6}$.