Counting and probability (Mathematics A)

The sum is 5 for (large, small) = (1,4), (2,3), (3,2) and (4,1), which is 4 ways. Since the dice are distinguishable, (1,4) and (4,1) are counted separately. Forgetting the distinction and counting only {1,4} and {2,3} as 2 ways is a common slip. The count 5 belongs to a sum of 6.

The sum is a multiple of 4 when it is one of 4,8,12. A sum of 4 arises in 3 ways as (1,3)(2,2)(3,1), a sum of 8 in 5 ways as (2,6)(3,5)(4,4)(5,3)(6,2), and a sum of 12 in 1 way as (6,6). These cannot happen together, so the addition principle gives 3+5+1=9 ways. Forgetting (6,6) leaves 8.

For each of the 3 shirts there are 4 choices of trousers, and for each of those 2 choices of shoes, so the multiplication principle gives 3×4×2=24 ways. Adding as 3+4+2=9 is a common slip, and 12 forgets the shoes.

There are 33 multiples of 3 and 20 multiples of 5, and 6 multiples of 15 lie in both. By n(A∪B)=n(A)+n(B)−n(A∩B) we get 33+20−6=47. Forgetting to subtract the overlap and writing 33+20=53 is the commonest mistake.

The product is even exactly when at least one die is even. Rather than counting directly, it is quicker to subtract the odd-product cases, both dice odd, which number 3×3=9, from the total of 36, giving 36−9=27 ways. The count 9 reports the complementary event itself, and 18 assumes without reason that it must be half.

The ways of using the coins number 2×3×4=24, taking 0 or 1 hundreds, 0 to 2 fifties and 0 to 3 tens, and removing the zero-yen case leaves 23. But since 100 yen equals two 50-yen coins, different ways of using the coins can produce the same amount. Writing out the amounts, the distinct values from 10 to 230 yen number 19. Confusing the 23 ways of using the coins with the number of amounts is a common slip.

Factorising as 720=2⁴×3²×5, a divisor is fixed by choosing the exponent of 2 in 5 ways, of 3 in 3 ways and of 5 in 2 ways, giving (4+1)(2+1)(1+1)=30. Forgetting to add 1 to each exponent and writing 4×3×2=24 is a common slip, and 29 excludes 1 from the divisors.

Each term comes from choosing one letter from each bracket and multiplying, so by the multiplication principle there are 2×3×2=12. Adding as 2+3+2=7 is a common slip, and 24 multiplies one factor too many, as in 2×3×2×2.

₅P₃ is the product of 3 integers counting down from 5: 5×4×3=60. The value 10 is ₅C₃, a combination, so take care not to mix permutations and combinations. The value 125 is 5³, which allows repetition.

The two posts are different, so the order of choosing matters and this is a permutation: ₁₀P₂=10×9=90 ways. Forgetting the distinction between the posts and writing ₁₀C₂=45 is the commonest mistake. The count 100 is 10×10, which wrongly allows the same person to hold both posts.

All 4 distinct students are arranged, so there are 4!=4×3×2×1=24 ways. The counts 16 as 4² and 256 as 4⁴ both come from assuming 4 choices at each position, which allows repetition.

Treating the two girls as one block, the 3 boys and the block make 4 items, arranged in 4!=24 ways. Inside the block the girls can be arranged in 2!=2 ways, so the answer is 24×2=48 ways. Forgetting the internal 2! and answering 24 is a common slip, and 120 is 5!, the count with no condition.

Fix the most restricted places first, the two ends. The left end can be any of 3 boys and the right end any of the remaining 2, giving ₃P₂=6 ways. The 3 people in the middle can be arranged in 3!=6 ways, so the multiplication principle gives 6×6=36 ways. Counting the ends correctly as ₃P₂=6 and then doubling again for swapping left and right gives 72. The count 12 forgets the 3!=6 arrangements in the middle and writes ₃P₂×2.

For a circular permutation, fix one person and arrange the other 4, giving (5−1)!=4!=24 ways. The count 5!=120 counts separately the 5 seatings that coincide under rotation.

Treating A and B as one block gives a circular permutation of 6 items, which is (6−1)!=120 ways. Multiplying by the 2! arrangements of A and B inside the block gives 120×2=240 ways. Forgetting the 2! leaves 120, and 720 is 6!, counting without the circular reduction.

This is a permutation with repetitions, so 5!/(2!2!)=120/4=30 ways. The count 5!=120 counts swaps of identical letters more than once, and 60 divides by 2! only once.

6!/(3!2!)=720/12=60 ways. Divide by both the 3! swaps of the letters a and the 2! swaps of the letters b. Dividing only by 3! gives 120, and only by 2! gives 360.

The hundreds digit cannot be 0, so there are 4 choices for it, then 4 remaining choices for the tens digit, which may be 0, and 3 for the units digit. The multiplication principle gives 4×4×3=48. The value ₅P₃=60 also counts arrangements with 0 in front, which are two-digit numbers. The value 36 would come from using only 3 choices for the tens digit, forgetting that 0 may go there, and 100 from allowing repeats as 4×5×5.

₇C₂=7×6/(2×1)=21. The value 42 is ₇P₂, forgetting to divide by 2!. Since a combination ignores order, divide the permutation count by r!, the number of rearrangements.

₈C₅=₈C₃=8×7×6/(3×2×1)=56. Choosing 5 is the same as leaving 3, so it is quicker to compute with the smaller r. The value 336 is 8×7×6 without dividing by 3!, and 28 is ₈C₂.

With no distinction of roles this is a combination: ₁₂C₃=12×11×10/(3×2×1)=220 ways. The count 1320 is ₁₂P₃, and mixing up permutations with combinations is the commonest mistake. When there is no need to arrange, divide by r!.

The boys can be chosen in ₅C₂=10 ways and the girls in ₄C₂=6 ways, so the multiplication principle gives 10×6=60 ways. The count 126 is ₉C₄, choosing 4 from all 9 without regard to sex, which ignores the condition.

Choosing 2 of the 8 vertices gives ₈C₂=28 segments, but the 8 sides joining adjacent vertices are not diagonals. So the answer is 28−8=20. Forgetting to subtract the sides and answering 28 is a common slip.

Choosing who goes into the group of 4 fixes the group of 5 automatically, so the answer is ₉C₄=126 ways. Groups of different sizes are automatically distinguishable, so you must not divide by 2!, which would give 63. The count 252 doubles instead.

Choosing in turn gives ₆C₂×₄C₂=15×6=90, but since the groups are not distinguished, each split has been counted once for each of the 3!=6 orderings of the groups. So the answer is 90÷3!=15 ways. Forgetting to divide by 3! and answering 90 is the commonest mistake.

Of the 7 moves, 4 east and 3 north, choose which 3 are north: ₇C₃=35 ways. Equivalently, count the arrangements of the string east-east-east-east-north-north-north as 7!/(4!3!)=35. Writing 4×3=12 is a wrong way to count, and 210 is ₇P₃, treating it as a permutation.

In total there are ₉C₄=126 routes. Routes through P number ₄C₂=6 from A to P times ₅C₂=10 from P to B, that is 60. So the routes avoiding P number 126−60=66. The count 60 answers the through case, and 126 is the total without the condition.

Regard A and B as already chosen and pick 2 more from the remaining 8: ₈C₂=28 ways. The count 56 is ₈C₃, taking 3 more by mistake, and 45 is ₁₀C₂, mishandling the condition.

There are 6×6=36 outcomes in total. The sum is 7 for (1,6)(2,5)(3,4)(4,3)(5,2)(6,1), which is 6 outcomes, so P=6/36=1/6. Treating (1,6) and (6,1) as the same gives 3 and hence 1/12, but since the denominator 36 counts the dice as distinct, the numerator must do so too.

There are ₅C₂=10 outcomes in total, and ₃C₂=3 ways to choose 2 red, so P=3/10. The value 9/25=(3/5)² applies when the ball is replaced, which is wrong here because drawing both at once means no replacement.

The complementary event, neither showing a 6, has probability (5/6)²=25/36, so P=1−25/36=11/36. The value 25/36 reports the complementary probability itself. The value 1/3=2/6 comes from casually adding because there are two places a 6 could appear, which counts (6,6) twice.

The complementary event is all three showing tails, with probability (1/2)³=1/8, so P=1−1/8=7/8. Whenever the phrase at least appears, think of the complement. The value 1/8 reports the complement itself.

There are 6 multiples of 3, 5 multiples of 4, and 1 multiple of 12 lying in both. So P=(6+5−1)/20=10/20=1/2. Forgetting to subtract the overlap, the multiple of 12, gives 11/20. For events that are not mutually exclusive, the addition rule always subtracts P(A∩B).

For independent trials the probabilities multiply: P=(2/3)×(3/5)=6/15=2/5. The value 13/15 is the probability that at least one hits and 2/15 that both miss, so take care to identify the event asked for.

The formula for repeated trials gives ₄C₂(1/6)²(5/6)²=6×(1/36)×(25/36)=25/216. Forgetting the factor ₄C₂ gives 25/1296, which is the commonest mistake, and forgetting (5/6)² gives 1/6. Never forget ₄C₂, the choice of which 2 throws show the 1.

₅C₃(1/2)³(1/2)²=10×(1/2)⁵=10/32=5/16. Forgetting the factor ₅C₃=10 gives 1/32. Even when heads and tails are equally likely, the choice ₅C₃ of which tosses show heads is still needed.

Once A has won, 9 lots remain of which 2 are winners, so the conditional probability is 2/9. The value 3/10 is B's unconditional probability of winning, which is also 3/10 for the second drawer by the fairness of lot drawing, and 1/15=(3/10)×(2/9) is the probability that both win, which is easily confused with the conditional probability.

A sum of at least 10 occurs in 6 ways: (4,6)(5,5)(6,4)(5,6)(6,5)(6,6). Of these, 5 contain a 6, all but (5,5). Narrowing the denominator to the 6 outcomes with sum at least 10 gives 5/6. The value 5/36 keeps the denominator at 36 and is the joint probability; for a conditional probability the denominator becomes the conditioning set.

E=100×(2/10)+50×(3/10)+0×(5/10)=20+15+0=35 yen. The value 50 is the plain average (100+50+0)/3 of the three prize amounts, ignoring the probability weights. An expected value is always the sum of value times probability.

The complementary event, no 6 in any of the 3 throws, has probability (5/6)³=125/216, so P=1−125/216=91/216. The value 125/216 reports the complement itself. The value 25/72=75/216 is the probability of exactly one 6, ₃C₁(1/6)(5/6)², so mind the difference from at least once. The value 1/2 comes from casually adding 1/6 three times.