Fundamental Counting Principle Worksheets - Math Monks - Free Printable

Explanation:
Let’s go through each problem one by one.

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Problem 1:
*There are 7 sprinters in a 100-yard race. What are the different possibilities of getting a top three finish?*

We care about order (1st, 2nd, 3rd), so this is a permutation problem.

We choose 3 sprinters out of 7, and order matters.

Number of ways = $ P(7,3) = 7 \times 6 \times 5 = 210 $

✔ Check:
- 7 choices for 1st place
- Then 6 left for 2nd
- Then 5 left for 3rd
→ $7 \cdot 6 \cdot 5 = 210$

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Problem 2:
*How many different 4-digit even numbers can be written with 0, 5, 6, 3, 8, 7?*

Digits available: {0, 3, 5, 6, 7, 8} — total 6 digits.
We need 4-digit even numbers → last digit must be even.

Even digits from the set: 0, 6, 8 → 3 choices for units digit.

But we must be careful: the first digit cannot be 0 (since it's a 4-digit number).

So we’ll split into cases based on what the last digit is.

Case A: Last digit = 0
Then remaining digits to choose for thousands, hundreds, tens places: from {3,5,6,7,8} → 5 digits left.
- Thousands place: can’t be 0 (but 0 is already used as last digit, so all 5 remaining are non-zero) → 5 choices
- Hundreds: 4 remaining
- Tens: 3 remaining
→ $5 \times 4 \times 3 = 60$ numbers

Case B: Last digit = 6
Used digit: 6
Remaining digits: {0,3,5,7,8} → 5 digits
- Thousands place: cannot be 0 → so only 4 choices (3,5,7,8)
- Hundreds: now 4 digits left (including 0)
- Tens: 3 left
→ $4 \times 4 \times 3 = 48$

Case C: Last digit = 8
Same as case B (symmetric): used 8, remaining {0,3,5,6,7}
- Thousands: 4 choices (not 0)
- Hundreds: 4
- Tens: 3
→ $4 \times 4 \times 3 = 48$

Total = 60 + 48 + 48 = 156

✔ Double-check:
Yes, that’s correct. Some might mistakenly do $3 \times 5 \times 4 \times 3 = 180$, but that ignores the restriction that first digit ≠ 0 when 0 is still available.

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Problem 3:
*A single die is rolled twice. In how many ways can you roll a number that is prime, followed by a 6?*

A standard die has faces: 1, 2, 3, 4, 5, 6
Prime numbers on a die: 2, 3, 5 → 3 primes
Second roll must be 6 → only 1 choice

So total favorable outcomes = 3 (choices for first roll) × 1 (second roll = 6) = 3

✔ Example outcomes: (2,6), (3,6), (5,6)

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Problem 4: Evaluate

a) $6!$
= $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

b) $\frac{5!}{2! \cdot 3!}$
First compute:
$5! = 120$, $2! = 2$, $3! = 6$
So denominator = $2 \cdot 6 = 12$
$\frac{120}{12} = 10$

This is actually $\binom{5}{2} = 10$ or $\binom{5}{3} = 10$

c) $\frac{6!}{2!}$
$6! = 720$, $2! = 2$
$\frac{720}{2} = 360$

d) $5C_2 = \binom{5}{2} = \frac{5!}{2!3!} = 10$ (same as part b)

e) $9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024$

Check:
9×8 = 72
72×7 = 504
504×6 = 3024 ✔

f) $12P_3 = \frac{12!}{9!} = 12 \times 11 \times 10 = 1320$

g) $\frac{21! \cdot 3!}{4!}$
Note: $4! = 24$, $3! = 6$, so $\frac{3!}{4!} = \frac{6}{24} = \frac{1}{4}$
So expression = $ \frac{21!}{4} $

But unless they want it simplified symbolically, maybe they expect numeric?
Wait — 21! is huge, so likely they want it expressed as $\frac{21!}{4}$, but let’s double-check:

Actually, $\frac{21! \cdot 3!}{4!} = 21! \cdot \frac{6}{24} = 21! \cdot \frac{1}{4} = \frac{21!}{4}$

Yes — that’s simplest exact form. Since 21! is divisible by 4 (21! includes factors 2 and 4), it's an integer, but we don’t need to compute full value.

But maybe they expect simplification like:
$$
\frac{21! \cdot 6}{24} = \frac{21!}{4}
$$
That’s fine.

However, sometimes in worksheets, they expect you to cancel before computing. Let’s see if any cancellation with factorial terms — no, 21! doesn’t simplify further with 4! except as above.

We’ll keep as $\frac{21!}{4}$, but maybe they want numeric? Unlikely — too big. So symbolic is okay.

Wait — check problem statement: “Evaluate”. In many high school contexts, if it’s huge, they accept simplified factorial form. But let’s see other parts: they computed small ones numerically. Maybe g) is a trick? Let me recompute:

$\frac{21! \cdot 3!}{4!} = \frac{21! \cdot 6}{24} = \frac{21!}{4}$ — yes.

But perhaps they meant $\frac{21!}{4!} \cdot 3!$ — same thing.

We’ll write $\frac{21!}{4}$, or if they want integer, note that 21! = 51,090,942,171,709,440,000
Divide by 4 = 12,772,735,542,927,360,000
But that’s messy. Given other answers are small, maybe it's a typo? Alternatively, maybe it's $\frac{21 \cdot 3!}{4!}$? No, it says $21!$.

Given context, I think they expect simplified exact form: $\frac{21!}{4}$

But let’s hold — maybe we misread: part g is “$\frac{21\,3!}{4!}$” — no, original says “$\frac{21\,3!}{4!}$”? Wait, in the image it's written as:
g) $\frac{21\ 3!}{4!}$ — could be ambiguous. But in LaTeX it's written as `\frac{21 3!}{4!}` which usually means $21 \times 3!$, not $21!$. Ah! That’s likely.

Let me re-express: In the user message, it's written:
"g) $\frac{21 3!}{4!}$"

In math notation, when you write "21 3!" without a multiplication sign, it often means $21 \times 3!$, especially since $21!$ would be written with exclamation after 21.

Yes! That makes more sense — because otherwise it's absurdly large for a worksheet.

So reinterpret:
g) $\frac{21 \cdot 3!}{4!}$

Then:
$3! = 6$, $4! = 24$
So numerator = $21 \cdot 6 = 126$
$\frac{126}{24} = \frac{21}{4} = 5.25$? But that’s not integer — unusual for counting problems.

Wait — maybe it's $\frac{21!}{3! \cdot 4!}$? No, original says “21 3!” — could be typeset issue.

Looking back at the user input:
"g) $\frac{21 3!}{4!}$"

In many printed worksheets, they write like “21 3!” meaning “21 choose 3”, i.e., $_{21}C_3$, but that would be written as $ \binom{21}{3} $ or $21C_3$. However, part d is $5C_2$, part h is $7C_2$, so maybe g) is miswritten.

Wait — the list is:
a) 6!
b) 5! / (2! 3!)
c) 6! / 2!
d) 5C2
e) 9P4
f) 12P3
g) 21 3! / 4!
h) 7C2

Hmm. Possibly g) is $\frac{21!}{3! \cdot 4!}$? But that’s unusual.

Alternative: maybe it's $\frac{21 \times 3!}{4!} = \frac{126}{24} = \frac{21}{4}$ — but fractional answer seems odd in this context.

Let me check if 21·3! / 4! simplifies:
= 21 × 6 / 24 = 126 / 24 = 21 / 4 = 5.25 — unlikely.

What if it's $_{21}P_3$? That would be 21×20×19 = 7980 — plausible.

But the notation is clearly “\frac{21 3!}{4!}”.

Given the pattern, and that all other answers are integers, I suspect it's a formatting error and should be:
$\frac{21!}{3! \cdot 4!}$ — but still huge.

Wait — look at part b: $\frac{5!}{2! 3!}$ = C(5,2)
Part g might be $\frac{21!}{3! \cdot 18!}$ = P(21,3) or C(21,3)? No.

Another idea: maybe “21 3!” means “21 choose 3”, i.e., $ \binom{21}{3} $, and the fraction bar is a mistake. But it has /4!.

Let me count the items: a to h — 8 parts. d is 5C2, h is 7C2, so g might be something else.

Given time, and that in many similar worksheets, a common problem is:
$\frac{21 \cdot 20 \cdot 19}{3!} = \binom{21}{3} = 1330$, but that’s not matching.

Wait — perhaps it's $\frac{21!}{4!}$ divided by $3!$? No.

Let me assume the most plausible: it's $ \frac{21 \times 3!}{4!} $, and accept 21/4 — but that feels wrong.

Alternatively, could “21 3!” mean “21 factorial times 3 factorial”? That’s what we did first → 21!·6 / 24 = 21!/4 — integer, huge.

Since the worksheet includes 6!, 9P4, etc, and those are manageable, maybe g is a trick to leave as $\frac{21!}{4}$, but I think for a student worksheet, they wouldn’t include 21!.

Let me search my memory: sometimes problems write “21 3!” to mean permutation: $_{21}P_3 = 21×20×19$, and the “3!” is not part of numerator — but the LaTeX shows \frac{21 3!}{4!}, so 3! is in numerator.

Given the ambiguity, and to stay safe, I will compute as written:
$ \frac{21 \cdot 3!}{4!} = \frac{21 \cdot 6}{24} = \frac{126}{24} = \frac{21}{4} = 5.25 $

But since all other answers are integers, I’m now convinced it's likely a typo, and intended to be:
$\frac{21!}{3! \cdot 4!}$ — still big.

Wait — look at part c: $\frac{6!}{2!}$ = 720/2 = 360
Part b: 120/(2·6)=10
So they are doing factorial simplifications.

Maybe g) is $\frac{21 \cdot 20 \cdot 19 \cdot 18}{4!}$? That would be C(21,4) = 5985.

But without clearer info, and since the user expects an answer, and in many standard sets, problem g is:
$\frac{21!}{3! \cdot 4!}$ is not standard.

Let me check numeric: If it's $\binom{21}{3} = \frac{21×20×19}{6} = 1330$
If it's $\frac{21×3!}{4!} = 5.25$ — reject.

Given the pattern, and that part d is 5C2 = 10, part h is 7C2 = 21, maybe g is 21C3 = 1330.

And the notation “21 3!” might be a misrender of “21C3”, where “C” looks like space+3!.

Yes! Very likely: the original had “21C3” but OCR or typing turned it into “21 3!”.

Because:
- d) 5C2
- g) likely 21C3
- h) 7C2

That fits perfectly.

So I will treat g) as $ \binom{21}{3} = \frac{21 \times 20 \times 19}{3 \times 2 \times 1} = \frac{7980}{6} = 1330 $

That is integer, reasonable, and matches format.

We’ll go with that.

h) $7C_2 = \binom{7}{2} = \frac{7 \cdot 6}{2} = 21$

Now summarize all answers:

1. 210
2. 156
3. 3
4a. 720
4b. 10
4c. 360
4d. 10
4e. 3024
4f. 1320
4g. 1330 (assuming 21C3)
4h. 21

Let me verify 21C3: 21×20×19 = 7980; ÷6 = 1330 ✔

Yes.

Final Answer:
1. 210
2. 156
3. 3
4a. 720
4b. 10
4c. 360
4d. 10
4e. 3024
4f. 1320
4g. 1330
4h. 21