✎ Work on paper. The checker checks only your final number, not your method. Commas in answers are fine.
QUESTION 01 / 30 Core · 2 min
Cancel before calculating
Count & simplify
Evaluate 72! 70! × 2!
01 Need a starting point?+ Write 72! as 72 × 71 × 70!. Stop expanding when you reach the factorial in the denominator.
02 Reveal worked solution+ FINAL ANSWER 2,556
The exclamation mark means factorial. For example, 4! = 4 × 3 × 2 × 1. It is multiplication, not subtraction. Expand only the part you need: 72! = 72 × 71 × 70! . The same non-zero factor 70! appears above and below, so it cancels: 72 × 71 × 70! 70! × 2! 72 × 71 2 Divide 72 by 2 first: 36 × 71 = 2,556 . You never needed the enormous value of 72!. Watch this trap 72! ÷ 70! is 72 × 71, not 2! or 72 ÷ 70.
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Original practice · not a past-paper question
QUESTION 02 / 30 Core · 3 min
An unknown in a permutation
Count & simplify
Find the integer n ≥ 3 if n P3 = 6 × n P2
01 Need a starting point?+ Expand the left side as n(n − 1)(n − 2). Both sides have the factor n(n − 1).
02 Reveal worked solution+ FINAL ANSWER 8
n P3 = n(n − 1)(n − 2)n P2 = n(n − 1)n(n − 1)(n − 2) = 6n(n − 1) .Because n ≥ 3, the factor n(n − 1) is not zero. Divide both sides by it: n − 2 = 6 . Therefore n = 8 . Check: 8 × 7 × 6 = 6 × 8 × 7 . Watch this trap Cancel common factors, not isolated terms inside a sum or difference.
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Original practice · not a past-paper question
QUESTION 03 / 30 Core · 1 min
Repetition is allowed
Count & simplify
Using digits 1–5, count three-digit numbers when reuse is permitted.
01 Need a starting point?+ Choosing a digit does not remove it. How many choices remain for each of the three places?
02 Reveal worked solution+ FINAL ANSWER 125
All five available digits are non-zero. The first position therefore has 5 choices. Reuse is allowed, so the second and third positions also each have 5 choices. Multiply the numbers of choices: 5 × 5 × 5 = 5³ = 125 . Watch this trap The factors do not decrease when repetition is allowed.
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QUESTION 04 / 30 Core · 2 min
A passcode can begin with zero
Count & simplify
A passcode has five positions. Use digits 0–9 without repeating a digit. A leading zero is allowed. How many passcodes are possible?
01 Need a starting point?+ This is a code, not a five-digit number. Zero is a valid first character.
02 Reveal worked solution+ FINAL ANSWER 30,240
The first position can contain any of 10 digits, including 0. No digit can repeat, so the remaining positions have 9, 8, 7 and 6 choices. 10 × 9 × 8 × 7 × 6 = 30,240 . For example, 01234 is a valid five-position passcode.Watch this trap Do not ban a leading zero when the question explicitly allows it.
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Original practice · not a past-paper question
QUESTION 05 / 30 Core · 2 min
A number cannot begin with zero
Count & simplify
Count four-digit numbers whose digits are all different.
01 Need a starting point?+ The thousands digit has 9 choices. After choosing it, zero is still available for the next position.
02 Reveal worked solution+ FINAL ANSWER 4,536
A four-digit number cannot start with 0. Its first digit has 9 choices: 1 through 9. One digit has been used. There are 9 of the original 10 digits left for the second position, including 0. The next two positions have 8 and 7 choices. 9 × 9 × 8 × 7 = 4,536 .Watch this trap The first two factors are 9 and 9—not 9 and 8—because zero becomes usable after the first position.
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QUESTION 06 / 30 Core · 2 min
Equal combinations
Count & simplify
Find the integer n ≥ 7 if n C4 = n C7
01 Need a starting point?+ For a fixed n, choosing r objects gives the same count as leaving out r objects: ⁿCᵣ = ⁿCₙ₋ᵣ.
02 Reveal worked solution+ FINAL ANSWER 11
For valid integer indices, n Ca = n Cb Here 4 and 7 are different, so use the second possibility: n = 4 + 7 = 11 . The logic: choosing 4 of 11 determines exactly which 7 are left out. Those are the same number of selections. Watch this trap Equal counts do not always mean the two lower indices are equal.
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Original practice · not a past-paper question
QUESTION 07 / 30 Core · 2 min
Make the last digit even
Arrange with restrictions
Count three-digit even numbers using {1,2,3,4,6,7}, without digit reuse.
01 Need a starting point?+ Choose the units digit first from 2, 4 and 6. Then count the unused digits.
02 Reveal worked solution+ FINAL ANSWER 60
An even number must end in an even digit. There are 3 choices here: 2, 4 or 6. After reserving that last digit, 5 digits remain for the hundreds position. None is zero. Then 4 digits remain for the tens position. 3 × 5 × 4 = 60 . We are multiplying counts of choices, not multiplying the digits of a number.Watch this trap The last digit uses up one of the available digits. It cannot be used again.
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QUESTION 08 / 30 Exam · 4 min
Even numbers when zero is available
Arrange with restrictions
Using only {0,1,2,3,4,5}, form four-digit even numbers without repeated digits. How many are possible?
01 Need a starting point?+ Split into two disjoint cases: the last digit is 0, or the last digit is 2 or 4.
02 Reveal worked solution+ FINAL ANSWER 156
Case 1: last digit 0. The first digit has 5 choices; the middle positions then have 4 and 3. Count: 1 × 5 × 4 × 3 = 60 .Case 2: last digit 2 or 4. There are 2 choices for the end. Of the 5 remaining digits, one is zero, so only 4 can start the number.After fixing both ends, 4 digits remain for the hundreds position and 3 for the tens position. Count: 2 × 4 × 4 × 3 = 96 . The cases cannot overlap, so add them: 60 + 96 = 156 . Watch this trap Using 3 identical cases for endings 0, 2 and 4 fails: the first-digit restrictions differ.
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Original practice · not a past-paper question
QUESTION 09 / 30 Core · 2 min
A word beginning with a vowel
Arrange with restrictions
Rearrange every letter of MONDAY. How many arrangements begin with a vowel?
01 Need a starting point?+ The vowels in MONDAY are O and A. Choose the first letter, then arrange the rest.
02 Reveal worked solution+ FINAL ANSWER 240
There are two choices for the first letter: O or A. Treat Y as a consonant in this question. After choosing the first letter, the other five distinct letters can be ordered in 5! = 120 ways. 2 × 5! = 2 × 120 = 240 . Arrangements need not be dictionary words.Watch this trap Only the first letter is restricted. The rest can appear in any order.
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QUESTION 10 / 30 Core · 3 min
Identical letters are not new arrangements
Arrange with restrictions
How many distinct six-letter arrangements use all the letters of BANANA?
01 Need a starting point?+ There are 3 identical As, 2 identical Ns and 1 B. Swapping equal letters does not make a new arrangement.
02 Reveal worked solution+ FINAL ANSWER 60
Temporarily label the letters A₁, A₂, A₃, N₁, N₂ and B. These six labelled objects have 6! arrangements. Remove the labels. The 3! orders of the As all look the same, as do the 2! orders of the Ns. Each genuinely different arrangement was counted 3! × 2! times. Divide out that overcount. 6! 3! × 2! 720 6 × 2 Watch this trap Reusing a letter freely is not the same as rearranging a fixed set of repeated letters.
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Original practice · not a past-paper question
QUESTION 11 / 30 Core · 3 min
Two people must be together
Arrange with restrictions
Seven different friends stand in a row. Anmol and Riya must stand next to each other. How many line-ups are possible?
01 Need a starting point?+ Treat the pair as one block. The block can internally be AR or RA.
02 Reveal worked solution+ FINAL ANSWER 1,440
Join Anmol and Riya into a temporary block. Together with the other 5 people, there are now 6 objects. Arrange these 6 objects in 6! ways. Inside the block, the pair has 2 orders: AR and RA. Multiply: 6! × 2! = 720 × 2 = 1,440 . Watch this trap The block can move anywhere in the row; it is not forced to the start.
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Original practice · not a past-paper question
QUESTION 12 / 30 Exam · 3 min
Two people must not be together
Arrange with restrictions
The same seven different friends stand in a row. This time Anmol and Riya must NOT be neighbours. Count the line-ups.
01 Need a starting point?+ Count every line-up, then remove those where the pair is adjacent.
02 Reveal worked solution+ FINAL ANSWER 3,600
Without restrictions, the seven people have 7! = 5,040 line-ups. Adjacent line-ups: treat the pair as one block, giving 6! × 2! = 1,440 . Every line-up is either adjacent or non-adjacent, never both. Required count: 5,040 − 1,440 = 3,600 . Watch this trap “Not together” here concerns one specified pair, not every possible pair of friends.
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Original practice · not a past-paper question
QUESTION 13 / 30 Exam · 3 min
All vowels form one block
Arrange with restrictions
Rearrange all letters of DAUGHTER with the vowels in one consecutive block. Count arrangements.
01 Need a starting point?+ The vowels A, U and E form one object. How many objects do you then arrange?
02 Reveal worked solution+ FINAL ANSWER 4,320
The three vowels are A, U, E. The five consonants are D, G, H, T, R; all eight letters are distinct. Treat the vowels as one block. The block plus the 5 consonants makes 6 objects, arranged in 6! ways. The 3 vowels can change order inside the block in 3! ways. 6! × 3! = 720 × 6 = 4,320 .Watch this trap Do not forget the arrangements inside the vowel block.
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QUESTION 14 / 30 Exam · 4 min
No two vowels are neighbours
Arrange with restrictions
Rearrange all eight letters of DAUGHTER so that no two vowels are next to each other. Count the arrangements.
01 Need a starting point?+ Arrange the 5 consonants first. They create 6 gaps, including the two ends. Put at most one vowel in each chosen gap.
02 Reveal worked solution+ FINAL ANSWER 14,400
The five distinct consonants can be arranged in 5! = 120 orders. A consonant row creates six spaces: _ D _ G _ H _ T _ R _ . The displayed consonant order is just one example. Choose 3 of the 6 spaces, then arrange A, U, E in them: 6 C3 × 3! = 20 × 6 = 120 Total: 5! × 6 C3 × 3! = 120 × 120 = 14,400 . Watch this trap Subtracting “all three vowels together” from 8! is NOT enough: it still allows two vowels to touch.
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Original practice · not a past-paper question
QUESTION 15 / 30 Exam · 3 min
An alternating row
Arrange with restrictions
Four different boys and four different girls stand in one row, alternating boy–girl throughout. How many arrangements are possible?
01 Need a starting point?+ There are two patterns: BGBGBGBG and GBGBGBGB. Arrange the people inside each pattern.
02 Reveal worked solution+ FINAL ANSWER 1,152
Alternation leaves exactly two position patterns: boys first, or girls first. For either pattern, the four boys fill their four positions in 4! ways, and the girls fill theirs in 4! ways. Add the two equal cases: 2 × 4! × 4! = 2 × 24 × 24 = 1,152 . Watch this trap Count both starting patterns. The people are distinct even when their group labels match.
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Original practice · not a past-paper question
QUESTION 16 / 30 Challenge · 5 min
Not all identical letters together
Arrange with restrictions
Count rearrangements of MISSISSIPPI where the four Is do not form one consecutive block.
01 Need a starting point?+ Count all distinct arrangements. Subtract arrangements of an IIII block with the remaining letters.
02 Reveal worked solution+ FINAL ANSWER 33,810
The 11 letters consist of M once, I four times, S four times and P twice. All distinct arrangements: 11! 4! × 4! × 2! Put all four Is into one IIII block. Now arrange 8 objects: that block, M, four Ss and two Ps. Blocked arrangements: 8! 4! × 2! Subtract: 34,650 − 840 = 33,810 . Two or three Is may still touch; only the four-I block is excluded. Watch this trap “The four Is are not all together” does not mean “no two Is are adjacent”.
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QUESTION 17 / 30 Core · 2 min
Select a group, not a line-up
Select without overcounting
Eight different volunteers are available. Choose a three-person team with no separate roles. How many teams are possible?
01 Need a starting point?+ ABC and CBA are the same team. Divide the ordered count by the 3! orders of each team.
02 Reveal worked solution+ FINAL ANSWER 56
Choosing in sequence gives 8 × 7 × 6 = 336 ordered selections. Each single team has been counted 3! = 6 times. For example, ABC, ACB, BAC, BCA, CAB and CBA all use the same three people. Therefore 8 C3 = 8 × 7 × 6 3 × 2 × 1 Watch this trap A team with no separate roles is a combination, not a permutation.
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Original practice · not a past-paper question
QUESTION 18 / 30 Exam · 3 min
A committee with a chairperson
Select without overcounting
From 12 people, count five-member committees with one designated chairperson.
01 Need a starting point?+ First choose the chairperson. Then choose the other four committee members without assigning them roles.
02 Reveal worked solution+ FINAL ANSWER 3,960
There are 12 choices for the chairperson. Choose 4 of the remaining 11 people: 11 C4 = 11 × 10 × 9 × 8 4 × 3 × 2 × 1 Multiply: 12 × 330 = 3,960 . Check another way: choose the team first, then its chairperson: 12 C5 × 5 = 792 × 5 = 3,960 Watch this trap Do not use ¹²P₅: only the chairperson has a special role; the other four members are interchangeable as a group.
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QUESTION 19 / 30 Core · 2 min
Exactly three from each group
Select without overcounting
Select three boys and three girls from five boys and four girls. Count the teams.
01 Need a starting point?+ Choose the boys and choose the girls. Both selections are needed for each team.
02 Reveal worked solution+ FINAL ANSWER 40
Choose the 3 boys in 5 C3 = 10 Choose the 3 girls in 4 C3 = 4 Each boys’ group can go with every girls’ group. Total: 10 × 4 = 40 . Watch this trap Do not add the two counts. A complete team needs both groups.
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QUESTION 20 / 30 Exam · 3 min
At least one of two named people
Select without overcounting
A four-person team is chosen from ten students, including Anmol and Riya. The team must contain at least one of these two. How many teams are possible?
01 Need a starting point?+ Subtract the teams containing neither person from the total number of teams.
02 Reveal worked solution+ FINAL ANSWER 140
There are 10 C4 = 210 Excluding both named students leaves 8 people, who give 8 C4 = 70 The allowed count is 210 − 70 = 140 . Why not select one named student first? Teams containing both would then be counted twice. Watch this trap “At least one” includes both. Do not accidentally count only exactly one.
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Original practice · not a past-paper question
QUESTION 21 / 30 Exam · 4 min
At least three women
Select without overcounting
A group has six women and four men, all distinct. How many five-person teams contain at least three women?
01 Need a starting point?+ List the possible team compositions: 3 women + 2 men, 4 + 1, or 5 + 0.
02 Reveal worked solution+ FINAL ANSWER 186
For exactly 3 women and 2 men: 6 C3 × 4 C2 = 20 × 6 = 120 For exactly 4 women and 1 man: 6 C4 × 4 C1 = 15 × 4 = 60 For exactly 5 women and no men: 6 C5 × 4 C0 = 6 × 1 = 6 These cases cannot overlap. Add: 120 + 60 + 6 = 186 . Watch this trap Choosing three women first, then any two remaining people, overcounts teams containing four or five women.
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Original practice · not a past-paper question
QUESTION 22 / 30 Core · 2 min
Compulsory choices
Select without overcounting
Choose five courses from nine; two specified courses are compulsory. Count selections.
01 Need a starting point?+ The two compulsory courses are already fixed. Choose only the remaining three courses.
02 Reveal worked solution+ FINAL ANSWER 35
Include the two compulsory courses. They introduce no choice. There are 7 other courses available and 3 places left in the programme. Required count: 7 C3 = 7 × 6 × 5 3 × 2 × 1 Watch this trap Do not choose 5 from 7; two of the five required places are already filled.
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QUESTION 23 / 30 Exam · 4 min
Exactly one ace
Select without overcounting
Count five-card hands containing exactly one ace from a standard 52-card deck.
01 Need a starting point?+ A standard deck has 4 aces and 48 non-aces. Choose one from the first group and four from the second.
02 Reveal worked solution+ FINAL ANSWER 778,320
Choose the single ace in 4 C1 = 4 The other four cards must be non-aces. Choose them in 48 C4 48 C4 = 48 × 47 × 46 × 45 4 × 3 × 2 × 1 Multiply: 4 × 194,580 = 778,320 . Card order is irrelevant in a hand. Watch this trap Choosing the other four cards from all remaining 51 cards could include extra aces.
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QUESTION 24 / 30 Exam · 4 min
One choice requires another
Select without overcounting
Eight candidates include A and B. Choose six. Selecting A requires selecting B. Count valid groups.
01 Need a starting point?+ The only forbidden groups contain A but not B. Groups containing B without A are allowed.
02 Reveal worked solution+ FINAL ANSWER 22
All six-person groups: 8 C6 = 28 For a forbidden group, include A and exclude B. Choose its other 5 members from the 6 remaining people: 6 C5 = 6 Valid groups: 28 − 6 = 22 . Alternative: A absent gives 7 C6 = 76 C4 = 15 Watch this trap “If A, then B” is not the same as “both or neither”. B alone is allowed.
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QUESTION 25 / 30 Exam · 4 min
A range and a divisibility rule
Mixed exam challenge
Count integers strictly between 6000 and 7000 that are divisible by 5 and have no repeated digit.
01 Need a starting point?+ The thousands digit is fixed. A multiple of 5 must end in 0 or 5.
02 Reveal worked solution+ FINAL ANSWER 112
Every valid number starts with 6. That uses the digit 6. The last digit must be 0 or 5: 2 choices. Both differ from 6. The middle positions have 8 and 7 choices, because the first and last digits have already been used. Total: 1 × 2 × 8 × 7 = 112 . The endpoints do not qualify, so no further subtraction is needed. Watch this trap Zero may appear in the middle. Only the first digit of a multi-digit number cannot be zero.
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QUESTION 26 / 30 Challenge · 4 min
Dictionary rank
Mixed exam challenge
Order every permutation of RACHIT alphabetically. What position does RACHIT occupy, counting from 1?
01 Need a starting point?+ Count all arrangements that come before RACHIT, then add one for RACHIT itself.
02 Reveal worked solution+ FINAL ANSWER 481
Put the letters in alphabetical order: A, C, H, I, R, T. They are all different. Before any word beginning with R come those beginning with A, C, H or I. Each starting letter allows 5! = 120 arrangements. That accounts for 4 × 120 = 480 earlier words. After fixing R, the remaining letters in RACHIT are A, C, H, I, T—already in alphabetical order. So RACHIT is the first word beginning with R. Rank: 480 + 1 = 481 . Watch this trap A rank starts at 1. The number of earlier words is not itself the rank.
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QUESTION 27 / 30 Challenge · 5 min
An exam with section restrictions
Mixed exam challenge
Select eight questions: Part I has five, Part II has seven. At least three must come from each part. Count selections.
01 Need a starting point?+ The only possible splits (Part I, Part II) are (3,5), (4,4) and (5,3).
02 Reveal worked solution+ FINAL ANSWER 420
You cannot choose fewer than 3 from either part, and Part I contains only 5 questions. For a 3 + 5 split: 5 C3 × 7 C5 = 10 × 21 = 210 For a 4 + 4 split: 5 C4 × 7 C4 = 5 × 35 = 175 For a 5 + 3 split: 5 C5 × 7 C3 = 1 × 35 = 35 Add the disjoint cases: 210 + 175 + 35 = 420 . The order in which the questions are attempted does not matter. Watch this trap Do not select 3 from each part first and then any 2 extras: the same final set can be created several ways.
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QUESTION 28 / 30 Exam · 3 min
When three points do not make a triangle
Mixed exam challenge
Ten distinct points lie in a plane. Exactly four lie on one line; there are no other collinear triples. How many triangles can have vertices among the points?
01 Need a starting point?+ Choose any three points, then remove the selections where all three are on the special line.
02 Reveal worked solution+ FINAL ANSWER 116
Any three non-collinear points determine exactly one triangle. All selections of three points: 10 C3 = 120 The four collinear points give 4 C3 = 4 Required count: 120 − 4 = 116 . Watch this trap Do not remove triples with only two points on the line. They can form valid triangles with a point off it.
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QUESTION 29 / 30 Challenge · 6 min
Three consecutive combinations
Mixed exam challenge
Given n Cr − 1 = 36, n Cr = 84, n Cr + 1 = 126r C2
01 Need a starting point?+ Take ratios of consecutive combinations. Cancel factorials to obtain (n − r + 1)/r and (n − r)/(r + 1).
02 Reveal worked solution+ FINAL ANSWER 3
For adjacent lower indices, factorial cancellation gives n Cr n Cr − 1 n − r + 1 r 84/36 = 7/3 . Hence 3(n − r + 1) = 7r , which becomes 3n + 3 = 10r . The next ratio gives n Cr + 1 n Cr n − r r + 1 2n = 5r + 3 . Double the first equation: 6n + 6 = 20r . Triple the second: 6n = 15r + 9 . Substituting gives 15r + 15 = 20r , so r = 3 . The question asks for r C2 3 C2 = 3 Watch this trap With consecutive combinations, dividing is usually easier than expanding three huge factorial expressions.
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QUESTION 30 / 30 Challenge · 5 min
A block with several repeated letters
Mixed exam challenge
Rearrange ASSASSINATION with all four Ss consecutive. How many distinct arrangements result?
01 Need a starting point?+ The SSSS block is a single object with only one internal arrangement. Count repeated letters outside the block too.
02 Reveal worked solution+ FINAL ANSWER 151,200
Count carefully: A appears 3 times, S 4 times, I twice, N twice, and T and O once each. Total: 13 letters. Join the four Ss into one block. There are now 13 − 4 + 1 = 10 objects to arrange. Among these objects, the As still repeat 3 times, and the Is and Ns each repeat twice. Required count: 10! 3! × 2! × 2! 3,628,800 24 There is no extra factor of 4!: exchanging identical Ss changes nothing. Watch this trap A block of distinct letters has internal permutations; a block of four identical Ss does not.
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