Work Life & Job Hunting · SPI Prep Lab
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Three books are chosen from 5 different books and lined up on a shelf from the left. How many arrangements are there?
You choose and then arrange, so this is a permutation: 5×4×3=60. Answering 10 gives the number of selections without regard to order, confusing permutations with combinations. Answering 125 treats the same book as reusable, a permutation with repetition. Answering 15 comes from 5×3.
In how many ways can all four people A, B, C and D be lined up in a row?
All four are arranged, so 4!=4×3×2×1=24. Answering 12 stops the calculation at 4×3. Answering 16 comes from 4×4. Answering 256 treats the same person as reusable, 4×4×4×4.
One item is chosen from each of 3 kinds of shirt, 4 kinds of trousers and 2 kinds of hat. How many combinations are there?
The choices are made one after another, so the product rule applies: 3×4×2=24. Answering 9 adds them, 3+4+2, but choices made in succession multiply. Answering 12 forgets the hats, 3×4, and answering 6 forgets the trousers, 3×2.
Six people sit at a circular table. Counting arrangements that match under rotation as one, how many seatings are there?
Fix one person's seat and arrange the remaining five: (6−1)!=5!=120. Answering 720 gives 6!, counting rotations of the same arrangement separately. Answering 360 gives 6!/2, which would also treat mirror images as the same. Answering 36 comes from 6×6.
There is one card each bearing 0, 1, 2, 3 and 4. Using 3 of them to make a three-digit number, how many numbers can be made?
The hundreds digit has 4 choices, excluding 0; the tens digit has the 4 that remain; the units digit has 3. 4×4×3=48. Answering 60 allows 0 at the front, 5×4×3, forgetting how 0 must be handled. Answering 24 counts 4×3×2. Answering 100 treats digits as reusable, 4×5×5.
There are 3 kinds of symbol. Lining up 4 of them in a row, with any symbol usable as often as you like, how many arrangements are there?
Each of the first to the fourth position can be chosen freely from 3, so 3×3×3×3=81. Answering 64 swaps the number of kinds and the number of positions, 4×4×4. Answering 12 comes from 3×4. Answering 24 is 4!, which contradicts itself by trying to make all 4 symbols different when only 3 kinds exist.
Three boys and two girls line up in a row. In how many arrangements are the two girls always next to each other?
Treat the two girls as one block and arrange 4 objects: 4!=24. The two girls can swap within the block, so 24×2!=48. Answering 24 forgets the final 2!. Answering 72 gives the count where the girls are not adjacent, confusing it with the complement. Answering 120 is 5!, ignoring the condition.
Three boys and three girls line up in a row. In how many arrangements do boys and girls alternate?
There are 2 cases according to whether a boy or a girl is at the front. In each, the three boys can be arranged in 3!=6 ways and the three girls in 3!=6 ways. 2×6×6=72. Answering 36 forgets to multiply by the 2 for the front position. Answering 144 multiplies by 2 twice. Answering 720 is 6!, ignoring the condition.
From 4 people, one chairperson and one vice-chairperson are chosen. How many ways are there?
The two roles are different, so order matters. There are 4 choices for chairperson and 3 remaining for vice-chairperson: 4×3=12. Answering 6 ignores the distinction between the roles and counts choices of 2 people, confusing permutations with combinations. Answering 16 lets one person hold both, 4×4. Answering 4 counts only the chairperson and forgets to multiply by the vice-chairperson.
Five people sit at a circular table. In how many seatings are two particular people next to each other? Arrangements that match under rotation count as one.
Treating the two as one block gives a circular arrangement of 4 objects: (4−1)!=6. The two within the block can be ordered in 2 ways, so 6×2=12. Answering 24 gives the unconstrained circular arrangement of five, 4!=24, forgetting to use the condition. Answering 6 forgets the final factor of 2. Answering 48 counts 2×4! without making it circular.
Six people are lined up in a row. In how many arrangements does A come before B?
The 6!=720 arrangements of six people split exactly in half between those with A in front and those with B in front: 720/2=360. Answering 720 ignores the condition. Answering 240 divides 720 by 3. Answering 120 gives 5!.
Seven people sit at a circular table. In how many seatings do three particular people sit in a row together? Arrangements that match under rotation count as one.
Bundling the three into one block gives a circular arrangement of 5 objects: (5−1)!=24. The three within the block can be ordered in 3!=6 ways, so 24×6=144. Answering 4320 multiplies the circular arrangement of seven, 720, by 3!, overlooking that bundling reduces the count of objects to 5. Answering 24 forgets to multiply by 3!. Answering 720 is the unconstrained circular arrangement of seven.
Three representatives are chosen from 8 people. How many ways are there?
The three representatives are not distinguished from one another, so this is a combination: 8×7×6/(3×2×1)=56. Answering 336 counts the order too, 8×7×6, which counts each set of three 3!=6 times over. Answering 112 doubles 56. Answering 24 comes from 8×3.
Two people are chosen from 10 for cleaning duty. How many ways are there?
The two on duty are not distinguished, so 10×9/2=45. Answering 90 gives 10×9 with order attached, forgetting to divide by 2. Answering 20 comes from 10×2 and 100 from 10×10.
From a group of 5 men and 4 women, 2 men and 2 women are chosen. How many ways are there?
The men can be chosen in 5×4/2=10 ways and the women in 4×3/2=6 ways. Both are decided together, so multiply: 10×6=60. Answering 16 adds them, 10+6. Answering 240 counts the order too, 20×12. Answering 126 ignores the split by sex and chooses 4 from 9.
Three red balls and two white balls are lined up in a row. Treating balls of the same colour as identical, how many arrangements are there?
Treating all five as distinct gives 5!=120, but the 3!=6 swaps among the red balls and the 2!=2 swaps among the white ones give the same arrangement, so divide: 120/(6×2)=10. Answering 120 skips the division; 20 divides only by 3!; 60 divides only by 2!.
Six letters in all, three A's, two B's and one C, are lined up in a row. How many arrangements are there?
6!/(3!×2!×1!)=720/12=60. Answering 720 counts swaps among identical letters separately, skipping the division. Answering 120 divides only by 3!. Answering 30 divides by an extra 2! as if there were two C's.
On a grid of streets you reach your destination by going 4 blocks east and 3 blocks north. How many routes are there without doubling back?
Out of 7 moves in total, deciding which 3 are northward fixes the route: 7×6×5/(3×2×1)=35. Answering 12 comes from 4×3. Answering 21 miscounts the northward moves as 2. Answering 70 miscounts the moves as 8.
Three toppings are chosen from 5 kinds. Order does not matter. How many ways are there?
5×4×3/(3×2×1)=10. Thinking of it as choosing the 2 kinds left out gives the same, 5×4/2=10. Answering 60 counts the order too, 5×4×3, confusing permutations with combinations. Answering 15 comes from 5×3 and 20 from 5×4.
Nine people are split into three groups of three. The groups are not distinguished from one another. How many ways are there?
Splitting into three distinguished groups gives 84×20×1=1680. Since the groups are not distinguished, each split is counted 3!=6 times over, once per ordering of the groups: 1680/6=280. Answering 1680 is the classic slip of forgetting to divide by 3!. Answering 840 divides by 2. Answering 84 gives only the count of choosing 3 from 9.
Six people are put into room A and room B, three in each. The rooms are distinguished. How many ways are there?
Choosing the three for room A fixes the rest for room B, so 6×5×4/(3×2×1)=20. Answering 10 gives the count when the rooms are not distinguished, dividing by 2 when you should not. Answering 40 doubles 20. Answering 720 gives 6!.
Eight people are split into two groups of four. The groups are not distinguished. How many ways are there?
Choosing 4 from 8 gives 70, but that counts each split twice, once with the chosen four and once with the other four. 70/2=35. Answering 70 is the classic slip of forgetting to divide by 2. Answering 28 confuses it with choosing 2 from 8. Answering 16 comes from 4×4.
Six people are split into three groups of two. The groups are not distinguished. How many ways are there?
For three distinguished groups it would be 15×6×1=90. Since the groups are not distinguished, divide by the 3!=6 orderings of the groups: 90/6=15. Answering 90 skips the division. Answering 45 divides by 2. Answering 30 divides by 3 rather than by 3!.
A die is rolled once. What is the probability that the number shown is 3 or more?
The numbers 3 or more are 3, 4, 5 and 6, that is 4 out of 6 in all: 4/6=2/3. Answering 1/2 reads it as 4 or more, 3/6. Answering 1/3 reads it as 5 or more, 2/6. Answering 5/6 reads it as 2 or more. Note that 'or more' includes the number itself.
Three coins are tossed at once. What is the probability that all three come up heads?
There are 2×2×2=8 ways for heads and tails to fall, and all three heads is 1 of them: 1/8. Answering 1/6 miscounts the cases as 6. Answering 1/4 is the value for two coins. Answering 3/8 is the probability of exactly two heads.
A bag holds 4 red balls and 6 white balls. After a good shake one ball is drawn. What is the probability that it is red?
Of the 10 balls in all, 4 are red: 4/10=2/5. Answering 3/5 gives the probability of drawing a white ball, mistaking which colour is asked about. Answering 1/5 miswrites 4/10 as 2/10. Answering 1/2 treats it as a choice between two colours, but the counts differ, so it is not an even split.
Two dice are thrown at once. What is the probability that the numbers total 7?
There are 6×6=36 outcomes. A total of 7 comes from 1 and 6, 2 and 5, 3 and 4, and their reverses, which is 6 outcomes: 6/36=1/6. Answering 5/36 is the probability of a total of 6. Answering 1/12 ignores order and counts 3 outcomes, 3/36. Answering 1/7 counts unordered pairs in both numerator and denominator, out of 21 pairs, but those 21 pairs are not equally likely, so it is wrong.
Four coins are tossed at once. What is the probability that at least one comes up heads?
The complementary event is all four tails, with probability 1/16. The probability sought is 1−1/16=15/16. Answering 1/16 is the classic slip of giving the complement and forgetting to subtract from 1. Answering 1/4 is the probability of exactly one head, 4/16. Answering 1/2 is the value for a single coin.
Two balls are drawn together from a bag holding 3 red balls and 5 white balls. What is the probability that both are white?
There are 28 ways to draw 2 from 8, and 10 ways for both to be white, drawing 2 from 5: 10/28=5/14. Answering 25/64 squares 5/8 as if each ball were replaced; when both are drawn together the denominator for the second drops to 7, so it is wrong. Answering 3/28 is the probability that both are red. Answering 25/28 would be the probability that at least one is white.
A bag holds 5 white balls and 3 red balls. Two are drawn together. What is the probability that at least one is red?
The complementary event is both white, 10/28=5/14, so 1−5/14=9/14. Answering 5/14 gives the complement and forgets to subtract from 1. Answering 15/28 is the probability of exactly one red and leaves out the case of two reds. Answering 3/28 is the probability that both are red.
A hits the target with probability 2/3 and B with probability 1/2, and their results do not affect each other. What is the probability that at least one of them hits?
The complementary event is that both miss, 1/3×1/2=1/6, so 1−1/6=5/6. Answering 1/3 is the probability that both hit, 2/3×1/2. Answering 1/6 gives the complement itself. Answering 1/2 is the probability that exactly one hits. Note also that simply adding the probabilities, 2/3+1/2, exceeds 1, which shows addition cannot be the method.
Of 10 lots, 2 are winners. A and B each draw one lot in that order, and lots are not returned. What is the probability that B draws a winner?
If A wins it is 2/10×1/9, and if A loses it is 8/10×2/9. Adding gives 2/90+16/90=18/90=1/5. The order of drawing makes no difference; it stays 2/10=1/5. Answering 2/9 looks only at the case where A loses. Answering 1/45 is the probability that both win. Answering 1/10 miscounts the winners as 1.
Two balls were drawn together from a bag of 3 red and 4 white balls, and it is known that at least one is red. What is the probability that both are red?
There are 21 ways to draw 2 from 7. Both white accounts for 6, so at least one red accounts for 21−6=15. Of those, both red accounts for 3: 3/15=1/5. Answering 1/7 ignores the condition and keeps 21 as the denominator, 3/21. Answering 5/7 is the probability that at least one is red. Answering 3/7 is the probability that the first is red.
Of 10 lots, one first prize pays 500 yen, three second prizes pay 100 yen each, and the remaining six are blanks paying 0 yen. What is the expected amount received from drawing one lot?
Adding 500×1, 100×3 and 0×6 gives 800 yen, and dividing by the 10 lots gives 80 yen. Answering 200 yen averages over the 4 winning lots only, 800/4, leaving the blanks out of the denominator. Answering 800 yen is the total prize money itself. Answering 50 yen looks only at the first prize, 500/10.
Three shots are fired with a hit probability of 1/3 each. What is the probability of exactly two hits out of the three?
There are 3 ways to choose which two shots hit, and each has probability 1/3×1/3×2/3=2/27, so 3×2/27=6/27=2/9. Answering 2/27 forgets to multiply by the 3 cases. Answering 1/27 is the probability that all three hit. Answering 4/9 is the probability of exactly one hit.
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