The number of arrangements of the letters of the word BANANA in which the two N's do not appear adja

Q: Wait, why do we divide by 3! and 2! in the total arrangements calculation?

Great question! In permutations, if we have identical items, swapping them doesn't create a new, distinct arrangement. Since we have three A's and two N's, we divide by 3! and 2! to 'remove' the overcounted, identical-looking arrangements. It's like saying, 'I don't care which A is in which spot, as long as they are all A's!'

Q: Why don't we divide by 2! for the N's when we treat them as a single unit (NN)?

When we group the two N's into a single block (NN), they are no longer individual, swappable items. They are locked together in that block. Since they are identical, swapping them inside the block doesn't change the block itself. Therefore, we don't need to divide by 2! for the N's anymore.

Q: Can we solve this using the Gap Method instead?

Absolutely! The Gap Method is a fantastic way to verify your answer. You would first arrange the non-N letters (B, A, A, A) and then place the two N's into the gaps created between them. It's a more robust method for complex problems, and it will lead you to the exact same result of 40!

Question

The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently is

Options:

(A) 40

(B) 60

(C) 80

(D) 100

Sigma Percentile · Accepted Answer

### Analyzing the Total Landscape

To begin, we must determine the total number of arrangements of the letters in the word BANANA. The word consists of 6 letters in total: three A's, two N's, and one B.

When dealing with permutations of a multiset, we use the formula:
$$
	ext{Total Arrangements} = \frac{n!}{p!q!r!}
$$

Here, $n=6$, $p=3$ (for the A's), and $q=2$ (for the N's). Substituting these values, we get:
$$
	ext{Total} = \frac{6!}{3! 	imes 2!} = \frac{720}{6 	imes 2} = \frac{720}{12} = 60
$$
There are exactly $60$ distinct ways to arrange the letters of BANANA.

### The Constraint: Applying the Complementary Principle

The problem requires that the two N's are never adjacent. Calculating this directly is difficult, so we employ the **Complementary Principle**. We will calculate the "forbidden" cases—where the N's are together—and subtract them from the total.

To calculate the forbidden cases, we treat the two N's as a single "super-letter" or block: $(NN)$. Our new set of units to arrange becomes: $(NN)$, $B$, $A$, $A$, $A$.

### The Calculation of Forbidden Cases

We now have $5$ units to arrange. Among these, the three A's are identical. The number of ways to arrange these units is:
$$
	ext{Forbidden Arrangements} = \frac{5!}{3!}
$$

Note that we do not divide by $2!$ for the N's because they are now locked inside the $(NN)$ block and act as a single entity. The calculation yields:
$$
	ext{Forbidden} = \frac{120}{6} = 20
$$
There are $20$ arrangements where the N's are stuck together.

### Final Calculation

We have $60$ total arrangements and $20$ forbidden arrangements. The number of valid arrangements where the N's are not adjacent is:
$$
	ext{Valid Arrangements} = 60 - 20 = 40
$$

By using the **Complementary Principle**, we have successfully navigated the constraint. The final answer is **40**.

Animated Solution for Mathematics - Permutations and Combinations: The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently is

Select Answer:

Visualized Solution

The Word: $B A N A N A$

Total Arrangements Formula

Calculating Total Permutations

The Complementary Principle

Grouping the $N$ s Together

Permutations with $N$ s Together

Final Subtraction

Key Takeaway

The Sigma Insight: Linear Permutations

Analyzing the Total Landscape

The Constraint

The Calculation of Forbidden Cases

Final Calculation

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