Statistics Flashcards: Using Permutations And Combinations In Probability

What this deck covers

This deck focuses on Using Permutations And Combinations In Probability, giving you a quick way to review the definitions, rules, and examples that matter most for Statistics.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: From $10$ students, how many ways can you choose a president, vice president, and secretary?

Answer: $P(10,3)=720$. Order matters since positions are distinct.

Flashcard 2: Compute the number of $4$-digit codes using digits $0$-$9$ with no repetition allowed.

Answer: $P(10,4)=10\cdot 9\cdot 8\cdot 7=5040$. 10 choices, then 9, then 8, then 7 (no repeats).

Flashcard 3: A $5$-digit code uses digits $0$-$9$ with no repetition. How many codes are possible?

Answer: $P(10,5)=30240$. $P(10,5) = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.

Flashcard 4: State the formula for permutations of $n$ items taken $r$ at a time, $P(n,r)$.

Answer: $P(n,r)=\frac{n!}{(n-r)!}$. Order matters, so divide $n!$ by $(n-r)!$ to remove unused items.

Flashcard 5: Two cards are drawn without replacement. What is $P(\text{both aces})$?

Answer: $\frac{\binom{4}{2}}{\binom{52}{2}}$. Choose 2 aces from 4 divided by all ways to choose 2 from 52.

Flashcard 6: Compute $P(8,3)$, the number of ordered selections of $3$ from $8$.

Answer: $P(8,3)=8\cdot 7\cdot 6=336$. First 3 factors of $8!$: $8\cdot 7\cdot 6$.

Flashcard 7: Find the probability that a $5$-card hand contains at least $1$ ace (from a $52$-card deck).

Answer: $1-\frac{C(48,5)}{C(52,5)}$. 1 minus probability of no aces (complement rule).

Flashcard 8: State the multiplication rule for counting outcomes in a multi-stage process.

Answer: Total outcomes $=a_1a_2\cdots a_k$. Each stage multiplies the number of choices at that stage.

Flashcard 9: State the formula for the combination of $n$ objects taken $r$ at a time, $C(n,r)$.

Answer: $C(n,r)=\frac{n!}{r!(n-r)!}$. Divides by $r!$ to remove order from permutations.

Flashcard 10: Find the probability of getting exactly $2$ heads in $3$ fair coin flips.

Answer: $\frac{C(3,2)}{2^3}=\frac{3}{8}$. Choose 2 heads positions from 3, divide by $2^3$ total outcomes.

Flashcard 11: Compute $C(8,3)$, the number of unordered selections of $3$ from $8$.

Answer: $C(8,3)=56$. $\frac{8!}{3!5!}=\frac{336}{6}=56$.

Flashcard 12: State the classical probability formula using equally likely outcomes.

Answer: $P(E)=\frac{\text{favorable}}{\text{total}}$. Assumes all outcomes are equally likely to occur.

Flashcard 13: What is the factorial definition of $n!$ for integer $n\ge 1$, and what is $0!$?

Answer: $n!=n(n-1)\cdots 2\cdot 1$ and $0!=1$. Product of integers down to 1; special case $0!=1$.

Flashcard 14: Identify whether order matters for a permutation versus a combination.

Answer: Permutation: order matters; combination: order does not. Key distinction: ABC ≠ BAC for permutations, but ABC = BAC for combinations.

Flashcard 15: State the formula for the number of distinct permutations of $n$ items with repeats $n_1,n_2,\dots,n_k$.

Answer: $\frac{n!}{n_1!n_2!\cdots n_k!}$. Divides $n!$ by factorials of repeat counts to eliminate duplicates.

Flashcard 16: Compute the number of ways to arrange $6$ distinct books on a shelf.

Answer: $6!=720$. All arrangements of 6 distinct items: $6!$.

Flashcard 17: From $10$ students, how many ways can you choose a $3$-person committee?

Answer: $\binom{10}{3}=120$. Order doesn't matter for committee selection.

Flashcard 18: What is the fundamental counting principle for $k$ stages with $n_1,\dots,n_k$ choices?

Answer: $n_1\cdot n_2\cdots n_k$. Multiply choices at each stage for total outcomes.

Flashcard 19: Compute the number of ways to select a president, vice president, and secretary from $10$ people.

Answer: $P(10,3)=720$. Order matters for distinct positions: $10\cdot 9\cdot 8$.

Flashcard 20: How many distinct arrangements of the letters in $\text{LEVEL}$ are there?

Answer: $\frac{5!}{2!2!}=30$. LEVEL has 5 letters with L and E each appearing twice.

Flashcard 21: What is the relationship between permutations and combinations: express $P(n,r)$ using $\binom{n}{r}$?

Answer: $P(n,r)=\binom{n}{r}r!$. Each combination has $r!$ permutations when order is considered.

Flashcard 22: State the addition rule for any two events $A$ and $B$.

Answer: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Adds probabilities but subtracts overlap to avoid double-counting.

Flashcard 23: A box has $6$ red and $4$ blue marbles. Two are drawn without replacement. What is $P(\text{one red and one blue})$?

Answer: $\frac{\binom{6}{1}\binom{4}{1}}{\binom{10}{2}}=\frac{8}{15}$. Ways to pick 1 red and 1 blue divided by all ways to pick 2.

Flashcard 24: Compute $P(7,3)$.

Answer: $210$. $P(7,3) = \frac{7!}{4!} = \frac{5040}{24} = 210$.

Flashcard 25: State the complement rule for probability.

Answer: $P(E^c)=1-P(E)$. The probability of not-E equals 1 minus the probability of E.

Flashcard 26: State the formula for combinations of $n$ items taken $r$ at a time, $\binom{n}{r}$.

Answer: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$. Order doesn't matter, so divide by $r!$ to remove duplicate arrangements.

Flashcard 27: Compute the number of distinct permutations of the letters in $\text{MISSISSIPPI}$.

Answer: $\frac{11!}{4!4!2!}=34650$. Divide by factorials of repeated letters: M(1), I(4), S(4), P(2).

Flashcard 28: State the probability rule for independent events $A$ and $B$.

Answer: $P(A\cap B)=P(A)P(B)$. For independent events, multiply their individual probabilities.

Flashcard 29: Find the probability that a $5$-card hand contains exactly $2$ aces (from a $52$-card deck).

Answer: $\frac{C(4,2)C(48,3)}{C(52,5)}$. Choose 2 aces and 3 non-aces, divide by all 5-card hands.

Flashcard 30: State the formula for the permutation of $n$ objects taken $r$ at a time, $P(n,r)$.

Answer: $P(n,r)=\frac{n!}{(n-r)!}$. Divides $n!$ by $(n-r)!$ to remove the unused $(n-r)$ objects.

Flashcard 31: Compute the number of $4$-digit codes using digits $0$-$9$ with repetition allowed.

Answer: $10^4=10000$. Each position has 10 choices independently.

Flashcard 32: Find the probability of drawing $2$ hearts in $2$ draws without replacement from a $52$-card deck.

Answer: $\frac{C(13,2)}{C(52,2)}=\frac{1}{17}$. Choose 2 from 13 hearts over choose 2 from 52 cards.

Flashcard 33: What is the probability formula using equally likely outcomes and counting?

Answer: $P(A)=\frac{\#A}{\#S}$. Favorable outcomes divided by total outcomes.

Flashcard 34: Compute $\binom{7}{3}$.

Answer: $35$. $\binom{7}{3} = \frac{7!}{3!4!} = \frac{5040}{6 \times 24} = 35$.

Flashcard 35: Find the probability of drawing $2$ aces in $2$ draws without replacement from a $52$-card deck.

Answer: $\frac{C(4,2)}{C(52,2)}=\frac{1}{221}$. Choose 2 from 4 aces over choose 2 from 52 cards.

Flashcard 36: Compute the number of ways to choose a committee of $3$ from $10$ people.

Answer: $C(10,3)=120$. Order doesn't matter for committee selection.

Flashcard 37: A $5$-digit code uses digits $0$-$9$ with repetition allowed. How many codes are possible?

Answer: $10^5$. Each of 5 positions can be any of 10 digits: $10 \times 10 \times 10 \times 10 \times 10$.

Flashcard 38: Find the probability a random $4$-digit code (digits $0$-$9$) has no repeated digits.

Answer: $\frac{P(10,4)}{10^4}=\frac{5040}{10000}=0.504$. No-repeat codes divided by all possible codes.

Flashcard 39: A $5$-card hand is drawn from a $52$-card deck. What is $P(\text{all hearts})$?

Answer: $\frac{\binom{13}{5}}{\binom{52}{5}}$. Choose 5 hearts from 13 divided by all ways to choose 5 from 52.