Is 3! + 4! = 7!?

Exercise 6.2 · Q2

3! + 4! = 7!

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Concept understanding — Factorial Arithmetic

Factorial Arithmetic — From Intuition to Precision

Imagine you have 3 different books you want to arrange on a shelf. How many different ways can you line them up? You could try listing them: Book A, B, C — or A, C, B — or B, A, C — and so on. If you actually count, you'll find 6 arrangements.

Where does that 6 come from? For the first position, you have 3 choices. Once you pick one, you have 2 choices left for the second position. Then only 1 choice remains for the last spot. So the total is $3 \times 2 \times 1 = 6$ .

That product — multiplying a whole number by every positive integer smaller than it, all the way down to 1 — is called a factorial. It's one of the most useful shortcuts in counting.

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

The symbol is an exclamation mark: $n!$ is read as "n factorial". It only makes sense for non-negative integers.

The first few values

$n$	$n!$	Why it matters
0	1	Special case (explained below)
1	1	Only one way to arrange one thing
2	2	Two ways: AB or BA
3	6	Three books, six arrangements
4	24	Four items, 24 arrangements
5	120	Grows fast — five books, 120 ways

Note

$0! = 1$ is not a guess — it's defined to make formulas work. There is exactly one way to arrange zero objects: do nothing. Also, many formulas like $n! = n \times (n-1)!$ would break at $n=1$ if $0!$ weren't 1.

The recursive nature

Factorials have a beautiful pattern: every factorial is the current number times the previous factorial.

5! = 5 \times 4!

4! = 4 \times 3!

3! = 3 \times 2!

2! = 2 \times 1!

1! = 1 \times 0! = 1 \times 1 = 1

This recursive definition is often how you'll compute factorials in problems: $n! = n \times (n-1)!$ , with the base case $0! = 1$ .

Why factorials explode so fast

Notice how quickly the numbers grow: $5! = 120$ , $6! = 720$ , $7! = 5040$ , $10! = 3,628,800$ . By $20!$ , you're already at 2.4 quintillion. This rapid growth is why factorials appear in probability (counting arrangements of decks of cards), combinatorics (choosing teams), and even in advanced mathematics like Taylor series.

Watch out

A common mistake: thinking $n!$ means $n$ multiplied by something else, like $n$ times some number. It's not — it's the product of all integers from $n$ down to 1. Also, factorials are not defined for negative numbers or fractions in basic arithmetic.

The core idea in one sentence

Factorial arithmetic is simply the arithmetic of these products — adding, subtracting, multiplying, and dividing expressions that contain factorials. The key skill is learning to cancel common factors when simplifying, especially in fractions like $\frac{10!}{7!}$ .

For example:

\frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720

You never need to fully expand both factorials — just write out the part that doesn't cancel.

That's the intuition: factorials count arrangements, grow fast, and simplify beautifully when you keep them as products rather than computing the full number.

Factorial Arithmetic is a building block of the NCERT Class 11 Mathematics chapter on Permutations and Combinations, and mastering it is essential before tackling factorial-based important questions in CBSE board exams. Searches like "factorial arithmetic definition, formula and examples" or "n! formula class 11 maths" reflect exactly the kind of foundational practice this concept supports, and it remains a quick-scoring warm-up topic in JEE Main counting problems.

Factorials grow explosively, not additively. Computing both sides shows $3! + 4! = 30$ while $7! = 5040$ , so the equation is false.

Why factorials don't add like ordinary numbers

When you see $3! + 4!$ and wonder if it equals $7!$ , you're testing whether factorials behave like exponents (where $2^3 \cdot 2^4 = 2^7$ ) or like simple addition. They don't. A factorial $n!$ means multiplying all integers from $1$ to $n$ , and this multiplication grows so rapidly that adding two factorials gives a result vastly smaller than the factorial of their sum.

The key insight: $7!$ includes the product $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7$ , which is astronomically larger than just adding $3!$ and $4!$ .

Computing each side

Left side: $3! + 4!$

Start with each factorial:

3! = 3 \times 2 \times 1 = 6

4! = 4 \times 3 \times 2 \times 1 = 24

Adding them:

3! + 4! = 6 + 24 = 30

Right side: $7!$

Now compute the factorial of $7$ :

7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1

Work through the multiplication:

7 \times 6 = 42

42 \times 5 = 210

210 \times 4 = 840

840 \times 3 = 2520

2520 \times 2 = 5040

5040 \times 1 = 5040

So $7! = 5040$ .

Comparison

We have $30$ on the left and $5040$ on the right. These are nowhere close: $7!$ is 168 times larger than $3! + 4!$ .

Watch out

A common misconception is that factorials might combine additively like $a + b = c \implies a! + b! = c!$ . This never holds for $a, b \geq 2$ because factorial growth is multiplicative and explosive.

Tip

To see why factorials grow so fast, notice that $7! = 7 \times 6 \times 5 \times 4!$ . Even if we had $4!$ on the left, multiplying it by $7 \times 6 \times 5 = 210$ to get $7!$ shows the enormous gap.

✓Final answer

3! + 4! \neq 7!

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$n$
›Reveal solutionSolution
$n!(31-n)!$ is minimized exactly where $\binom{31}{n}$ is maximized — the middle of the range $0\le n\le31$ — giving $15!\,16!$ .

Concept and Intuition

Since $\binom{31}{n}=\dfrac{31!}{n!(31-n)!}$ and $31!$ is fixed, minimizing $n!(31-n)!$ is the same as maximizing $\binom{31}{n}$ . Binomial coefficients $\binom{N}{n}$ are known to be largest at the centre of the range $n=0,\dots,N$ and to decrease symmetrically toward the ends — so the minimum of $n!(31-n)!$ occurs at the centre, not at the extremes.

Step-by-Step Solution

We want to minimize $f(n)=n!(31-n)!$ for integer $n$ in $[0,31]$ .

Note $f(n) = \dfrac{31!}{\binom{31}{n}}$ , so minimizing $f(n)$ is equivalent to maximizing $\binom{31}{n}$ .

For $\binom{N}{n}$ , the maximum occurs at $n=\lfloor N/2\rfloor$ (and also at $n=\lceil N/2\rceil$ if $N$ is odd, by the symmetry $\binom{N}{n}=\binom{N}{N-n}$ ).

Here $N=31$ is odd, so the two largest, equal binomial coefficients occur at $n=15$ and $n=16$ : $\binom{31}{15}=\binom{31}{16}$ .

So the minimum of $f(n)=n!(31-n)!$ occurs at $n=15$ (or equivalently $n=16$ ): $f(15)=15!\times16!$ .

Common Mistakes

Assuming the minimum occurs at an endpoint ( $n=0$ or $n=31$ , giving $31!$ or $0!\,31!=31!$ ) — that's actually the maximum of $n!(31-n)!$ , not the minimum, since factorials grow so fast that splitting evenly makes the product of factorials smallest.

Forgetting the odd/even distinction: for odd $N=31$ the minimizing $n$ isn't unique (both 15 and 16 work), giving $15!\,16!$ (not a single fractional-index answer).

✓Final answer
$15!\,16!$

ANSWER: B
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$14$
›Reveal solutionSolution
The exponent of a composite number $6=2\cdot3$ in $n!$ is limited by whichever prime factor is scarcer — here, the exponent of 3, since 3's exponent (34) is much smaller than 2's (70).

Concept and Intuition

When a number is not prime, its exponent (the highest power dividing $n!$ ) is governed by the least available exponent among its prime factors, since you can only form as many complete '6's as you can pair a 2 with a 3, and 3s are always scarcer than 2s in $n!$ for $n\ge 2$ . Legendre's formula gives the exponent of a prime $p$ in $n!$ as $\sum_{k\ge1}\left\lfloor \dfrac{n}{p^k}\right\rfloor$ .

Step-by-Step Solution

Exponent of 2 in $72!$ : $\left\lfloor\frac{72}{2}\right\rfloor+\left\lfloor\frac{72}{4}\right\rfloor+\left\lfloor\frac{72}{8}\right\rfloor+\left\lfloor\frac{72}{16}\right\rfloor+\left\lfloor\frac{72}{32}\right\rfloor+\left\lfloor\frac{72}{64}\right\rfloor = 36+18+9+4+2+1=70$ .

Exponent of 3 in $72!$ : $\left\lfloor\frac{72}{3}\right\rfloor+\left\lfloor\frac{72}{9}\right\rfloor+\left\lfloor\frac{72}{27}\right\rfloor+\left\lfloor\frac{72}{81}\right\rfloor = 24+8+2+0=34$ .

Exponent of 6 in $72!$ $=\min(70,34)=34$ .

Common Mistakes

Adding the two exponents (70+34) instead of taking the minimum.

Forgetting a higher power term (e.g. $72/81=0$ ) or miscounting one of the floor divisions.

✓Final answer
$34$

ANSWER: C
AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
${}^{n}P_4 = 5040$
›Reveal solutionSolution
Match $^nP_4=5040$ to $n=10$ and $^{15}P_r=2730$ to $r=3$ by recognizing the products of consecutive integers, then compute $^{10}P_3=720$ .

Concept and Intuition

$^nP_k$ is just the product of $k$ consecutive integers counting down from $n$ . Rather than solving factorial equations algebraically, it's much faster to recognize these products directly by testing nearby integers, since permutation values grow fast and are easy to bracket.

Step-by-Step Solution

$^nP_4 = n(n-1)(n-2)(n-3) = 5040$ . Testing $n=10$ : $10\times9\times8\times7 = 5040$ . ✓ So $n=10$ .

$^{15}P_r = 15\times14\times\cdots\times(16-r) = 2730$ . Testing $r=3$ : $15\times14\times13 = 2730$ . ✓ So $r=3$ .

Now compute $^{n}P_r = {}^{10}P_3 = 10\times9\times8 = 720$ .

Common Mistakes

Forgetting $^nP_k$ has exactly $k$ factors, e.g. mis-testing $r=2$ ( $15\times14=210\ne2730$ ) or $r=4$ ( $15\times14\times13\times12=32760\ne2730$ ).

Confusing $n$ and $r$ when finally computing $^nP_r$ — it's $n=10$ from the first equation and $r=3$ from the second.

✓Final answer
$The correct option is (B) — 720.$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$^{27}P_{r+7} = 7722\ ^{25}P_{(r+4)}$
›Reveal solutionSolution
Write both permutation expressions using $^nP_k=\dfrac{n!}{(n-k)!}$ , simplify the factorial ratio, and solve the resulting linear equation for $r$ . Answer: $r=10$ .

Concept and Intuition

Permutation expressions with a variable in both the upper and lower index often simplify beautifully once you write everything as factorial ratios — most of the factorial cancels, leaving a simple linear (or low-degree) equation.

Step-by-Step Solution

$^{27}P_{r+7}=\dfrac{27!}{(27-(r+7))!}=\dfrac{27!}{(20-r)!}$ .

$^{25}P_{r+4}=\dfrac{25!}{(25-(r+4))!}=\dfrac{25!}{(21-r)!}$ .

The given equation: $\dfrac{27!}{(20-r)!}=7722\cdot\dfrac{25!}{(21-r)!}$ .

Write $27!=27\times26\times25!$ , so $\dfrac{27!}{25!}=702$ . The equation becomes $702\cdot\dfrac{25!}{(20-r)!}=7722\cdot\dfrac{25!}{(21-r)!}$ , and $25!$ cancels.

So $\dfrac{702}{(20-r)!}=\dfrac{7722}{(21-r)!}$ , i.e. $\dfrac{(21-r)!}{(20-r)!}=\dfrac{7722}{702}=11$ .

Since $(21-r)!=(21-r)\cdot(20-r)!$ , this ratio is simply $(21-r)$ . So $21-r=11\Rightarrow r=10$ .

Check validity: for $^{25}P_{r+4}$ to make sense we need $r+4\le25$ , i.e. $r\le21$ — satisfied by $r=10$ .

Common Mistakes

Miscomputing the factorial ratio $27!/25!$ (it's $27\times26=702$ , not $27$ alone).

Sign/index slip when expanding $(27-(r+7))$ or $(25-(r+4))$ .

✓Final answer
$10$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$7!$
›Reveal solutionSolution
Prime-factorizing $7!=2^4\cdot3^2\cdot5\cdot7$ and applying the divisor-count formula $(e_1+1)(e_2+1)\cdots$ gives $60$ divisors.

Concept and Intuition

If $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ , the total number of positive divisors of $n$ is $(e_1+1)(e_2+1)\cdots(e_k+1)$ , since each divisor is formed by independently choosing an exponent from $0$ to $e_i$ for each prime.

Step-by-Step Solution

$7! = 1\times2\times3\times4\times5\times6\times7 = 5040$ .

Prime factorize by counting powers of each prime $\le7$ among $1,\dots,7$ :

Power of $2$ : from $2,4=2^2,6=2\times3$ → $2^{1+2+1}=2^4$ .

Power of $3$ : from $3, 6=2\times3$ → $3^{1+1}=3^2$ .

Power of $5$ : from $5$ → $5^1$ .

Power of $7$ : from $7$ → $7^1$ . So $7! = 2^4\cdot3^2\cdot5^1\cdot7^1$ (check: $16\times9\times5\times7 = 144\times35=5040$ ✓).

Number of divisors $=(4+1)(2+1)(1+1)(1+1) = 5\times3\times2\times2$ .

$5\times3=15$ , $15\times2=30$ , $30\times2=60$ .

Common Mistakes

Forgetting that a divisor count formula uses (exponent + 1) for each prime, not the exponent itself.

Miscounting the power of $2$ or $3$ in the factorial (easy to under-count contributions from composite numbers like $4$ and $6$ ).

✓Final answer
$60$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$\displaystyle\sum_{r=1}^{15} r^2\left(\dfrac{{}^{15}C_r}{{}^{15}C_{r-1}}\right)=$
›Reveal solutionSolution
Simplify the binomial-coefficient ratio first, turning the sum into a simple polynomial sum; the total is $680$ .

Concept and Intuition

The ratio $\dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{n-r+1}{r}$ is a standard simplification that converts an awkward combinatorial sum into ordinary arithmetic series.

Step-by-Step Solution

With $n=15$ : $\dfrac{{}^{15}C_r}{{}^{15}C_{r-1}}=\dfrac{15-r+1}{r}=\dfrac{16-r}{r}$ .

So each term $r^2\cdot\dfrac{16-r}{r}=r(16-r)=16r-r^2$ .

Sum from $r=1$ to $15$ : $\displaystyle\sum 16r-\sum r^2=16\cdot\frac{15\cdot16}{2}-\frac{15\cdot16\cdot31}{6}$ .

$=16\times120-1240=1920-1240=680$ .

Common Mistakes

Misremembering the ratio formula (e.g. inverting it to $\frac{r}{16-r}$ ).

Arithmetic slips in $\sum r^2=\frac{n(n+1)(2n+1)}{6}$ for $n=15$ .

✓Final answer
$680$

ANSWER: B
AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$30^r$
›Reveal solutionSolution
$30^r=2^r3^r5^r$ divides $30!$ only as far as the scarcest prime factor allows — and $5$ appears only $7$ times in $30!$ , so $r=7$ .

Concept and Intuition

Legendre's formula gives the exact power of a prime $p$ dividing $n!$ : $\sum_{k\ge1}\lfloor n/p^k\rfloor$ . Since $30=2\times3\times5$ , $30^r$ needs $r$ copies of each of $2,3,5$ simultaneously; the answer is the minimum of the three individual prime-power counts in $30!$ .

Step-by-Step Solution

Power of $5$ in $30!$ : $\lfloor30/5\rfloor+\lfloor30/25\rfloor=6+1=7$ .

Power of $3$ in $30!$ : $\lfloor30/3\rfloor+\lfloor30/9\rfloor+\lfloor30/27\rfloor=10+3+1=14$ .

Power of $2$ in $30!$ : $\lfloor30/2\rfloor+\lfloor30/4\rfloor+\lfloor30/8\rfloor+\lfloor30/16\rfloor=15+7+3+1=26$ .

$30^r=2^r\cdot3^r\cdot5^r$ divides $30!$ only while $r\le\min(26,14,7)=7$ .

Common Mistakes

Forgetting to break $30$ into its prime factors and instead trying to count "how many multiples of 30" are in $30!$ directly (also works but is more error-prone).

Missing higher powers of $5$ or $3$ (e.g. forgetting $\lfloor30/25\rfloor=1$ ).

✓Final answer
$7$

ANSWER: B
AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$\dfrac{9}{8}(10)!$
›Reveal solutionSolution
Total arrangements minus the "E's glued together" arrangements gives the count with the two E's never adjacent: $\frac{9}{16}\cdot10!$ .

Concept and Intuition

The standard technique for "certain letters must NOT be adjacent" is: (Total arrangements) $-$ (arrangements where they ARE adjacent, found by gluing them into a single block). Both counts must account for the word's repeated letters correctly.

Step-by-Step Solution

ARRANGEMENT = A,R,R,A,N,G,E,M,E,N,T — 11 letters with A:2, R:2, N:2, E:2, and G,M,T:1 each.

Total distinct arrangements $=\dfrac{11!}{2!\,2!\,2!\,2!}=\dfrac{11!}{16}$ .

To count arrangements with the two E's adjacent, glue them into a single unit "EE" (they're identical, so no internal ordering needed). Now we arrange $10$ units: A,A,R,R,N,N,G,M,T,[EE] — repeats A:2,R:2,N:2 remain. Arrangements $=\dfrac{10!}{2!\,2!\,2!}=\dfrac{10!}{8}$ .

Arrangements with E's not adjacent $=\dfrac{11!}{16}-\dfrac{10!}{8}$ . Since $11!=11\times10!$ : $=\dfrac{11\times10!}{16}-\dfrac{2\times10!}{16}=\dfrac{9\times10!}{16}$ .

Common Mistakes

Forgetting one of the four repeated-letter pairs (A, R, N, or E) when dividing by factorials.

Adding instead of subtracting the "glued" count.

✓Final answer
$\dfrac{9}{16}(10)!$

ANSWER: C
AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$2^n$
›Reveal solutionSolution
Legendre's formula sums $\lfloor 16/2^i\rfloor$ for $i=1,2,3,4$ to get the exact power of $2$ in $16!$ , which is $15$ .

Concept and Intuition

The exponent of a prime $p$ in $n!$ is given by Legendre's formula: $\sum_{i=1}^{\infty}\left\lfloor\dfrac{n}{p^i}\right\rfloor$ — this counts multiples of $p$ , then multiples of $p^2$ (which contribute an extra factor), and so on, until $p^i>n$ .

Step-by-Step Solution

Apply Legendre's formula with $n=16$ , $p=2$ : sum $\left\lfloor\frac{16}{2}\right\rfloor + \left\lfloor\frac{16}{4}\right\rfloor + \left\lfloor\frac{16}{8}\right\rfloor + \left\lfloor\frac{16}{16}\right\rfloor + \left\lfloor\frac{16}{32}\right\rfloor+\ldots$

$\left\lfloor\frac{16}{2}\right\rfloor=8$ , $\left\lfloor\frac{16}{4}\right\rfloor=4$ , $\left\lfloor\frac{16}{8}\right\rfloor=2$ , $\left\lfloor\frac{16}{16}\right\rfloor=1$ , and $\left\lfloor\frac{16}{32}\right\rfloor=0$ (terms vanish beyond this).

Sum: $8+4+2+1=15$ .

So $2^{15}$ divides $16!$ exactly (i.e. $2^{15}\mid16!$ but $2^{16}\nmid16!$ ), giving $n=15$ .

Common Mistakes

Stopping the sum too early (e.g. only counting $\lfloor16/2\rfloor+\lfloor16/4\rfloor$ ) and missing the higher-power contributions from $8=2^3$ and $16=2^4$ .

✓Final answer
$15$

ANSWER: B
AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$f(n) = n!(31-n)!$
›Reveal solutionSolution
$f(n)=n!(31-n)!$ is inversely proportional to $\binom{31}{n}$ , so it's smallest where the binomial coefficient is largest — at $n=15$ or $16$ , giving minimum value $15!\cdot16!$ .

Concept and Intuition

Binomial coefficients $\binom{N}{n}$ are largest at the middle value(s) of $n$ and taper off symmetrically toward the ends. Since $f(n)=n!(31-n)!=31!/\binom{31}{n}$ , maximizing $\binom{31}{n}$ minimizes $f(n)$ .

Step-by-Step Solution

Note $\binom{31}{n}=\dfrac{31!}{n!(31-n)!}=\dfrac{31!}{f(n)}$ , so $f(n)=\dfrac{31!}{\binom{31}{n}}$ .

$f(n)$ is minimized exactly when $\binom{31}{n}$ is maximized.

Since $31$ is odd, $\binom{31}{n}$ attains its maximum at the two central, equal values $n=15$ and $n=16$ (as $\binom{31}{15}=\binom{31}{16}$ ).

So the minimum value of $f(n)$ is $f(15)=15!\,(31-15)!=15!\,16!$ (equal to $f(16)=16!\,15!$ ).

Common Mistakes

Assuming the minimum occurs at the extremes ( $n=0$ or $31$ ) — those actually give the maximum of $f(n)$ (since $\binom{31}{0}=1$ is the minimum binomial coefficient).

Picking only one of $15,16$ without checking they give equal, and correct, values.

✓Final answer
$(15!)(16!)$

ANSWER: D
AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$N(n)=n\displaystyle\prod_{r=1}^{2023}(n^2-r^2)\ (n>2023)$
›Reveal solutionSolution
The product telescopes to $N(2024)=4047!$ , and since $\binom{N}{N-1}=N$ for any $N$ , the answer is simply $4047!$ .

Concept and Intuition

Two ideas combine here: (1) the product $\prod_{r=1}^{k}(n-r)(n+r)$ telescopes into ratios of factorials, and (2) the combinatorial identity $\binom{N}{N-1}=N$ (choosing $N-1$ out of $N$ is the same as leaving one out, of which there are $N$ ways) means we never actually need to compute a huge binomial coefficient — just $N$ itself.

Step-by-Step Solution

$N(n) = n\displaystyle\prod_{r=1}^{2023}(n^2-r^2) = n\displaystyle\prod_{r=1}^{2023}(n-r)(n+r)$ .

At $n=2024$ : as $r$ runs $1$ to $2023$ , $(n-r)$ runs from $2023$ down to $1$ , so $\prod(n-r)=2023!$ .

As $r$ runs $1$ to $2023$ , $(n+r)$ runs from $2025$ to $4047$ , so $\prod(n+r)=\dfrac{4047!}{2024!}$ (product of all integers up to 4047, divided by the product up to 2024).

So $N(2024) = 2024\times2023!\times\dfrac{4047!}{2024!}$ . Since $2024\times2023!=2024!$ , this simplifies to $\dfrac{2024!}{2024!}\times4047!=4047!$ .

Finally, $\binom{N}{N-1} = \dfrac{N!}{(N-1)!\,1!} = N = 4047!$ .

Common Mistakes

Trying to actually expand or estimate the binomial coefficient $\binom{N}{N-1}$ as if it were a complicated quantity, missing that it always simply equals $N$ .

✓Final answer
$(4047)!$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$n$
›Reveal solutionSolution
Using Legendre's formula for the power of 5 dividing $n!$ , only $n=4009$ (among the given choices) gives exactly 1000 trailing zeros; the count jumps to 1001 at $n=4010$ and sits at 999 for $n=4000$ or $4004$ .

Concept and Intuition

The number of trailing zeros in $n!$ equals the exponent of $5$ in its prime factorisation (since factors of $2$ are always in surplus), computed by Legendre's formula: $Z(n)=\sum_{i=1}^{\infty}\left\lfloor \dfrac{n}{5^i}\right\rfloor$ . This function only increases at multiples of $5$ , and stays constant over each run of 5 consecutive integers between one multiple of 5 and the next.

Step-by-Step Solution

Compute $Z(4009) = \left\lfloor\frac{4009}{5}\right\rfloor+\left\lfloor\frac{4009}{25}\right\rfloor+\left\lfloor\frac{4009}{125}\right\rfloor+\left\lfloor\frac{4009}{625}\right\rfloor+\left\lfloor\frac{4009}{3125}\right\rfloor$ $=801+160+32+6+1 = 1000$ .

Compute $Z(4010) = 802+160+32+6+1 = 1001$ (jumps because $4010$ is again a multiple of 5, adding one more to the first term).

Compute $Z(4000) = 800+160+32+6+1=999$ and $Z(4004)=800+160+32+6+1=999$ (both below $4005$ , the next multiple of 5 where the count steps up to 1000).

So the count is exactly 1000 for $n=4005,4006,4007,4008,4009$ , and only $4009$ appears among the answer choices.

Common Mistakes

Forgetting the higher-power terms ( $\lfloor n/625\rfloor$ , $\lfloor n/3125\rfloor$ ) which still contribute at this magnitude of $n$ .

Assuming zeros increase steadily by exactly 1 at every integer, rather than remaining constant over blocks of 5 and jumping by more than 1 at multiples of 25, 125, etc.

✓Final answer
$4009$

ANSWER: C

←Previous question · 6.3.2Q1 — Evaluate (i) 8! (ii) 4! - 3!→Next question · 6.3.2Q3 — Compute (8!)/(6!×2!)

Q.Is 3!+4!=7!3! + 4! = 7!3!+4!=7!?

Why factorials don't add like ordinary numbers

Computing each side

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Q.Is $3! + 4! = 7!$ ?

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