Repair a factorial ratio without dropping the factorials

A factorial ratio changes when its factorial symbols are dropped before division. This original paper exercise begins with 6!/4!, then tries to replace it with 6/4. Expanding the two factorials gives 720/24, which equals thirty. The unsupported shortcut gives three halves. The repair identifies the actual shared product and keeps the factors that remain after it is removed.

The inputs here are exact nonnegative integers, with no measured quantities or recorded learner results. Define 0! = 1. For a positive integer n, define n! as the product of the integers from one through n. These definitions supply every factorial used below. An exclamation mark in this exercise names that product; it is part of the mathematical input.

Preserve the attempted shortcut

Write 6!/4! as the starting expression and retain 6/4 beside it as the attempted replacement. The fraction bar divides the complete numerator product by the complete denominator product. A note saying only “cancel the factorials” would leave unclear what values were supposedly divided out. Keeping the two actual expressions makes that proposed change available for an exact check.

The factorial marks do not supply an independent common number that can be erased from both sides. By definition, 6! stands for one through six multiplied together, while 4! stands for one through four multiplied together. Those complete products have a shared portion. We need to expose that portion before deciding which factors survive the division.

The distinction matters even though six and four remain printed in both lines. The numerator of the original fraction is 720, and its denominator is twenty-four. The numerator and denominator of the attempted replacement are six and four. Removing the marks has changed both quantities, and it has not yet shown that their quotient stays the same.

Expand the original products

Write the numerator as 1 times 2 times 3 times 4 times 5 times 6. Write the denominator as 1 times 2 times 3 times 4. Multiplying the numerator successively gives 2, 6, 24, 120 and 720. Multiplying the denominator gives twenty-four. The original fraction therefore becomes 720/24, retaining the full products supplied by its notation.

Twenty-four times thirty equals 720, so the original quotient is thirty. As a separate calculation, the attempted 6/4 reduces to 3/2. Multiplying 3/2 by twenty-four gives thirty-six, not 720. This reverse check shows that three halves cannot be the quotient of the original numerator by its original denominator. All comparisons are exact fractions and integers.

Keep that failed check attached to the proposed replacement. Six divided by four really is three halves; that final division has not gone wrong. The unsupported step was substituting six and four for their factorial products. A repair that merely changes the last answer to thirty would lose the explanation of where the value changed.

Cancel the complete shared factor

The numerator can be grouped as (1 times 2 times 3 times 4) times (5 times 6). The denominator is the first grouped product, which equals 4!. Thus the original ratio is [4! times (5 times 6)]/4!. The quantity being divided out is now explicit. It is the nonzero number twenty-four, present as a factor of both whole quantities.

Dividing numerator and denominator by that common factor leaves 5 times 6 above a denominator of one. The retained quotient is thirty, agreeing with the direct 720/24 evaluation. Every numerator factor has an identified place: one through four belong to the removed shared product, while five and six remain. No numerator factor has disappeared merely because of its punctuation.

The result contains two consecutive factors, not just six. The denominator stopped at four, so the numerator continues beyond its shared part with five and then six. Keeping only the final numeral six would omit five. Keeping six divided by four would also restore a denominator that was already wholly removed with the common product. Neither form records the expansion we actually performed.

You may write the surviving factors as six times five instead of five times six. Their product is still thirty. That rearrangement changes the written order of multiplication without changing the retained factors. It gives a compact form that will match the general expression n times (n minus one), while the expanded one-through-six version remains available to check the cancellation.

Check the next smaller inputs separately

For a separate control, use 5!/3!. The full numerator is 120 and the denominator is six, giving twenty. Expanding the numerator as 3! times four times five leaves the consecutive factors four and five after the common 3! is divided out. Dropping the marks would instead produce 5/3, a different exact fraction.

The control 4!/2! gives twenty-four divided by two, or twelve. Its shared product is two factorial, leaving three times four. The unsupported replacement 4/2 gives two. These controls preserve the same two-step difference between the factorial arguments. They are additional inputs under the same stated rule, rather than changes made to rescue the original six-over-four calculation.

At 3!/1!, the denominator product is one and the numerator is six. The retained factors are two and three, so the quotient is six. Dropping the marks gives 3/1 = 3. A denominator equal to one therefore does not make the proposed deletion valid. The factorial in the numerator still names its complete product, including the factor two.

Retain the zero-factorial boundary

The next input is 2!/0!. Our definition makes the denominator one, so this original ratio is defined and equals two. The expression obtained by dropping the marks would be 2/0, which has a zero denominator. That attempted replacement is undefined. The boundary comparison needs this explicit notation because the factorial argument zero and the factorial value one are different quantities.

For this boundary, the surviving product is still two times one. There is no need to write a descending multiplication that continues through zero. The definition 0! = 1 supplies the denominator directly. Keep that declared value beside the boundary calculation, so an empty product is not mistaken for multiplication by zero or silently removed from the domain statement.

The family considered here starts at n = 2. Smaller nonnegative values of n would make n minus two negative, while our factorial definition covers only nonnegative integers. We therefore make no evaluation of n!/(n minus two)! for those inputs. Restricting the family preserves the denominator definition used in the proof that follows.

Derive the consecutive-factor identity

Let n be an integer at least two. The product defining n! contains the product through n minus two, followed by n minus one and n. Consequently n! equals (n minus two)! times (n minus one) times n. At n = 2 this statement reads 2! = 0! times one times two, which also follows from our boundary definition.

Every factorial in the denominator is positive: 0! equals one, and a product of positive integers is positive. We may therefore divide both the numerator and denominator by (n minus two)!. The ratio becomes n times (n minus one). This is the same operation used for the common 4! in the original example, with its shared product stated for the whole family.

Substituting n = 6 gives six times five, or thirty. Substituting n = 5 gives five times four, or twenty. These checks connect the general form to the expanded examples without claiming that a few numerical cases alone prove every input. The product definition and the positive common denominator supply the general argument; the substitutions check how the notation is being used.

Test whether the shortcut can coincide

For integers n of at least three, dropping the marks would produce n divided by (n minus two). Both that denominator and n are positive. If this shortcut equalled the correct n times (n minus one), division by n and multiplication by n minus two would require (n minus one) times (n minus two) to equal one.

For n of at least three, n minus one is at least two and n minus two is at least one. Their product is therefore at least two, so it cannot equal one. There is no agreeing input in this integer range. The remaining boundary n = 2 has already failed differently because the dropped-mark expression has a zero denominator.

This conclusion is bounded to the ratio whose factorial arguments differ by two and to the integer domain we defined. It is not a rule about arbitrary punctuation, a count of real arrangements or an instruction for a calculator. Other factorial expressions need their own expansions. The current repair retains the original two products and the precise shared portion that they contain.

Check the retained product without a long expansion

The general form also lets us check a larger exact input without multiplying out its entire factorial. For 8!/6!, the numerator is 6! times seven times eight. Dividing by the same positive 6! leaves fifty-six. As a reverse check, six factorial is 720, and fifty-six times 720 is 40,320, the product of the integers through eight. The original denominator and the retained quotient reconstruct the complete numerator.

Dropping the marks in that input would give 8/6 = 4/3. Multiplying four thirds by the original denominator 720 gives 960, which does not reconstruct 40,320. The reverse check uses the same denominator for both proposed answers. Comparing each answer only with its own altered denominator would verify two separate divisions while failing to test whether the replacement preserved the original ratio.

For the original six-over-four input, that check is thirty times twenty-four equals 720. The rejected three-halves answer returns thirty-six when multiplied by twenty-four. Keep the common denominator fixed during this comparison so that the written repair tests the expression that was actually assigned.

Read the actual yo-yo accounts separately

The TOP YO Encores account repeats its title in the description. Its one image shows a complete pink yo-yo printed Encores and TOPYOnation. The native signature spelling and smaller pictured lettering remain separate observations. No material, supplied-string or performance claim is independently verified here.

The General Yo Royal 1 account identifies SKU sku-772712828 and UPC 634547839122. Three images show the blue primary toy, Royal 1 and GENERAL-YO markings, and its central connecting hardware. No independent spare part is established as included. Neither account supplies the factorial exercise or a tested learning outcome.

Keep neighboring mathematical decisions distinct

The additive-term fraction repair rejects deletion of matching addends across separate sums. This factorial quotient already consists of products. Expanding its notation identifies a real common product and the consecutive factors left after that complete factor is divided out.

The four-card derangement count classifies eligible orders by their cycle types. No arrangement count or cycle classification supplies the quotient here. The present decision is which factors remain when an explicitly defined numerator product is divided by its shorter shared product.

For the original input, keep 6!/4! = 720/24 = 6 times 5 = 30 in the answer field. Retain 6/4 = 3/2 as the rejected replacement, with the dropped factorial marks identified as its unsupported step. The correct cancellation removes the whole shared 4! and leaves both consecutive numerator factors visible.

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