Words of length 10 are formed using the letters, A, B, C, D, E, F, G, H, I, J. Let x be the number o

Q: Wait, why do we divide by $2!$ in the calculation for $y$?

Great question! When all letters are distinct, there are $10!$ ways to arrange them. However, when two letters are identical, swapping them doesn't create a new, unique word. Dividing by $2!$ (which is $2$) effectively 'removes' these redundant, identical-looking arrangements, ensuring we only count unique words.

Q: Why is it $\binom{9}{8}$ for the remaining letters?

We have $10$ total letters available. We already 'used' one letter to be the repeated pair. That leaves us with $9$ distinct letters. We need to fill $8$ more slots, and since no other letter can be repeated, we must choose $8$ distinct letters from those $9$ available. Hence, $\binom{9}{8}$.

Q: Can I solve this using the 'slot method' instead of combinations?

You absolutely can! You could pick the two slots for the repeated letter in $\binom{10}{2}$ ways, then fill the remaining $8$ slots with the remaining $9$ letters in $P(9, 8)$ ways. If you calculate it that way, you will arrive at the exact same result for $y$. Combinatorics is beautiful because there are often multiple paths to the same truth!

Question

Words of length 10 are formed using the letters, A, B, C, D, E, F, G, H, I, J. Let $x$ be the number of such words where no letter is repeated; and let $y$ be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, $\frac{y}{9x} =$

Sigma Percentile · Accepted Answer

### The Art of Counting: A Combinatorial Journey

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey into the heart of combinatorics.

We have ten letters—$A$ through $J$—and we are tasked with building words of length ten. It sounds simple, but within these constraints lies a beautiful dance of selection and arrangement.

### Phase 1: The World of $x$ (The Perfect Order)

Let us first define $x$. This is the number of words where no letter is repeated.

Imagine standing before ten empty slots, holding ten distinct letters. For the first slot, you have ten choices. For the second, nine remain. For the third, eight, and so on, until the final slot is filled by the last remaining letter.

This is the classic definition of a permutation of ten distinct items, which we elegantly write as $10!$. So, $x = 10!$. This is our baseline, our 'perfect' scenario where every letter is unique.

### Phase 2: The World of $y$ (The Slight Imperfection)

Now, let us turn our attention to $y$. Here, the rules change. We are told that exactly one letter is repeated twice, and no other letter is repeated.

In the first act, we must choose the 'star' of our word—the letter that gets the privilege of appearing twice. We have ten letters to choose from, so there are $\binom{10}{1} = 10$ ways to make this selection.

In the second act, we need to fill the remaining eight slots. We have used one letter for our pair, leaving us with nine distinct letters. We must choose eight of them to fill the remaining positions. This is $\binom{9}{8}$, which is simply $9$.

In the third act, we have our ten letters in hand: one identical pair and eight unique letters. If all ten were distinct, we would have $10!$ arrangements.

Because two letters are identical, swapping them does not produce a new word. To correct for this overcounting, we must divide by $2!$. Thus, the number of arrangements is:

$$ \frac{10!}{2!} $$

Putting it all together, our total count for $y$ is the product of these acts:

$$ y = \binom{10}{1} 	imes \binom{9}{8} 	imes \frac{10!}{2!} $$

Simplifying this, we get:

$$ y = 10 	imes 9 	imes \frac{10!}{2} $$

### Phase 3: The Grand Finale

We are now at the finish line. The problem asks for the ratio $\frac{y}{9x}$. Let us substitute our hard-won values into this expression:

$$ \frac{y}{9x} = \frac{10 	imes 9 	imes \frac{10!}{2}}{9 	imes 10!} $$

Look at the elegance of the cancellation! The $10!$ in the numerator and denominator vanish into thin air. The $9$ in the numerator and denominator also cancel out perfectly.

We are left with a simple, clean expression: $\frac{10}{2}$.

And there it is: **$5$**.

It is easy to get lost in the factorial notation or the binomial coefficients, but when you break the problem down into these logical steps—selection, curation, and symmetry correction—the complexity dissolves. You have mastered the logic of permutations.

Enter Numerical Value:

Visualized Solution

Understanding the Given Elements

Defining Condition for $x$

Calculating $x$

Defining Condition for $y$

Step 1: Choosing the Repeated Letter

Step 2: Choosing Remaining Letters

Step 3: Arranging the Letters

Formulating $y$

Simplifying $y$

Setting up the Ratio $\frac{y}{9 x}$

Simplifying the Fraction

Final Calculation

The Sigma Insight: Combinations and Selection

The Art of Counting

Phase 1

Phase 2

Phase 3

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