◆ QA · Modern Math

Probability , formulas + CAT PYQs

Focused Modern Math kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Probability is here.

1CAT PYQs

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Modern Math, formula sheet

Show the full Modern Math formula sheet (explanations + basic examples)

1Rule of Product & Rule of Sum

In plain English: when steps happen one after another you multiply; when you pick just one of several separate options you add.
Product (AND): if a job has m ways for step 1 and n ways for step 2 done together, total = m × n.
Sum (OR): if an action has m ways and a mutually-exclusive action has n ways, choose one in m + n ways.
e.g. 3 shirts AND 2 trousers → 3 × 2 = 6 outfits; but "a shirt OR a trouser" → 3 + 2 = 5 single items.

AND → × · OR → +

2Permutations (order matters)

In plain English: count the ways to line up r things in a row out of n, where swapping the order makes a new arrangement.
Arrangements of r out of n distinct things.
n! = n × (n−1) × … × 2 × 1 ; 0! = 1.
e.g. gold/silver/bronze from 5 runners: ⁵P₃ = 5 × 4 × 3 = 60.

nPr = n! / (n − r)!

3Combinations (order ignored)

In plain English: count the ways to pick a group of r things out of n when the order you pick them in does not matter.
Selections of r out of n distinct things.
Symmetry: choosing r = leaving out (n−r).
e.g. a team of 2 from 5 people: ⁵C₂ = (5 × 4)/(2 × 1) = 10.

nCr = n! / [(n − r)! r!] = nPr / r! nCr = nC(n−r)

4Sum of all combinations

In plain English: each item is either "in" or "out," so the total number of possible selections (including picking nothing) is 2 multiplied by itself n times.
Total subsets of an n-element set (incl. empty & full).
"Choose some-or-none" of n distinct items = 2ⁿ.
e.g. toppings from 3 choices: 2³ = 8 possible orders (incl. a plain pizza with none).

nC0 + nC1 + nC2 + … + nCn = 2ⁿ

5Arrangements with repetition (alike things)

In plain English: when some items are identical, swapping them changes nothing, so you shrink the plain n! by dividing out those wasted swaps.
n things where p are alike of one kind, q of another, r of a third.
Divide n! by the factorials of each repeated group.
e.g. arrangements of the letters of "LEVEL" (5 letters, L×2, E×2): 5!/(2! 2!) = 120/4 = 30.

Arrangements = n! / (p! q! r! …)

6Selecting from "some or all" of mixed items

In plain English: for each type you decide "how many to take" (0 up to all of them), multiply those choices, then knock off the one case where you took nothing.
p of one type, q of a second, r of a third, … (alike within a type).
Take any number (incl. none) of each, then subtract the all-empty case.
e.g. select some fruit from 2 alike apples & 3 alike oranges: (2+1)(3+1) − 1 = 12 − 1 = 11 ways.

{(p+1)(q+1)(r+1)…} − 1

7Circular arrangements

In plain English: around a circle there is no "first seat," so rotating everyone gives the same arrangement, fix one person and arrange the rest in a line.
Fix one person to kill rotational duplicates.
If clockwise = anticlockwise (e.g. a necklace), divide by 2.
e.g. 5 people at a round table: (5 − 1)! = 4! = 24 ways.

Round table = (n − 1)! · Necklace = (n − 1)!/2

8Dividing into groups

In plain English: choosing who goes in the first group automatically fixes the rest; if the groups have no labels and are the same size, swapping the two groups is a duplicate so divide by 2.
Split (m + n) things into two labelled groups of m and n.
If the two groups are equal (m = n), divide by 2! for identical groups.
e.g. split 6 people into groups of 4 and 2: 6!/(4! 2!) = 720/48 = 15.

(m + n)! / (m! n!) Equal groups: (2m)! / [2! (m!)²]

9Distributing identical things (partitions)

In plain English: line up the n identical items as "stars" and slot in (r − 1) dividers ("bars") to split them among the r people, count where the bars go.
Distribute n identical items among r persons, each may get any number (incl. 0).
"Stars & bars." For "each gets ≥ 1", first give one to each then apply the formula on what remains.
e.g. give 5 identical chocolates to 3 kids (any can get 0): (5 + 3 − 1)C(3 − 1) = ⁷C₂ = 21.

Ways = (n + r − 1)C(r − 1)

10Shortest grid paths

In plain English: a shortest path is just a sequence of "rights" and "ups"; count how many ways to order those moves.
To go across a grid using only two directions (m of one, n of the other).
Equivalent to arranging m + n moves of two kinds.
e.g. corner to corner of a 2×3 block (2 ups, 3 rights): (2 + 3)!/(2! 3!) = 120/12 = 10 paths.

(m + n)! / (m! n!) = (m+n)Cm

11Distinct terms in a multinomial expansion

In plain English: every term looks like aˣbʸcᶻ with x+y+z = n; counting the distinct terms is the same as counting whole-number ways to split n among three slots.
Number of terms in (a + b + c)ⁿ = non-negative integer solutions of a + b + c = n.
e.g. terms in (a + b + c)²: (2 + 2)C2 = ⁴C₂ = 6 (namely a², b², c², ab, bc, ca).

Terms in (a+b+c)ⁿ = (n + 2)C2

12Probability, definition

In plain English: probability is just "how many ways it can happen" divided by "how many ways anything can happen," assuming every outcome is equally likely.
Assumes equally likely outcomes; always between 0 and 1.
e.g. rolling an even number on a die: 3 favourable / 6 total = 1/2.

P(E) = favourable outcomes / total outcomes P(not E) = 1 − P(E)

13Odds

In plain English: odds pit the "wins" directly against the "losses" as a ratio, instead of dividing wins by the total like probability does.
Compares favourable to unfavourable cases (not to the total).
e.g. for rolling a 6 on a die: odds in favour = 1 : 5 (one good face, five bad).

Odds in favour = favourable : unfavourable Odds against = unfavourable : favourable

14Probability, addition law

In plain English: for "A or B," add the two chances but subtract the part you counted twice (where both happen).
General OR rule (subtract the overlap).
Mutually exclusive ⇒ P(A ∩ B) = 0.
e.g. a card that is a King or a Heart: 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Mutually exclusive: P(A or B) = P(A) + P(B)

15Probability, multiplication law

In plain English: for "A and B" of unrelated events, multiply their chances; if one affects the other, use the conditional version.
For independent events the AND probability multiplies.
Conditional probability links them when not independent.
e.g. two heads in two coin tosses: 1/2 × 1/2 = 1/4.

Independent: P(A ∩ B) = P(A) × P(B) Conditional: P(A | B) = P(A ∩ B) / P(B)

16Expected value

In plain English: weight each possible payoff by how likely it is, then add them up, what you'd average if you repeated it forever.
The long-run average of a random quantity.
Sum of (each value × its probability).
e.g. win ₹10 on heads, ₹0 on tails: E = 10 × ½ + 0 × ½ = ₹5.

E(X) = Σ xᵢ · P(xᵢ)

17Set theory, basics

In plain English: sets are just collections; "union" pools everyone, "intersection" keeps only the shared members, "difference" removes the overlap.
Union ∪ = in A or B (or both); Intersection ∩ = in both.
Difference A − B = in A but not B; Complement A′ = not in A.
Null set ⌀ is a subset of every set; power set of n elements has 2ⁿ subsets.
e.g. A = {1,2,3}, B = {2,3,4}: A∪B = {1,2,3,4}, A∩B = {2,3}, A−B = {1}.

A − B = {x : x ∈ A and x ∉ B}

18Two-set Venn formula

In plain English: people who do both got counted in each circle, so add the two totals and subtract that overlap once.
Add the two sets, subtract the double-counted overlap.
e.g. 30 like tea, 25 like coffee, 10 like both → total who like at least one = 30 + 25 − 10 = 45.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

19Three-set Venn formula

In plain English: add all three circles, take out each pairwise overlap (over-subtracting the centre), then add the triple-overlap back once to fix it.
Inclusion-exclusion: add singles, subtract pairs, add back the triple.
e.g. |A|=|B|=|C|=10, each pair shares 3, all three share 1 → union = 30 − 9 + 1 = 22.

n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)

20Venn, "exactly" layers (CAT favourite)

In plain English: someone in exactly two sets is counted twice in the size-total, someone in all three is counted thrice, these two equations untangle the layers fast.
Let x = exactly-one, y = exactly-two, z = exactly-three (all three).
Total in at least one set = x + y + z; the "repeated total" of memberships = sum of the set sizes.
e.g. sizes sum to 50, at least-one T = 30, all-three z = 5 → 30 + y + 2(5) = 50 → exactly-two y = 10.

x + y + z = T x + 2y + 3z = n(A) + n(B) + n(C)

21Arithmetic Progression (AP)

In plain English: an AP adds the same fixed step each time; the sum is just the average of the first and last term, times how many terms.
Constant common difference d; nth term and sum below.
e.g. 2, 5, 8, 11, 14 (a=2, d=3): 5th term = 2 + 4×3 = 14; sum = 5/2 × (2 + 14) = 40.

aₙ = a + (n − 1)d Sₙ = n/2 · [2a + (n − 1)d] = n/2 · (first + last)

22Geometric Progression (GP)

In plain English: a GP multiplies by the same fixed ratio each time; if that ratio is a fraction the terms shrink and the infinite sum settles on a finite value.
Constant ratio r; sum of a finite GP and (for |r|<1) an infinite GP.
e.g. 1 + ½ + ¼ + ⅛ + … (a=1, r=½): S∞ = 1/(1 − ½) = 2.

aₙ = a·r^(n−1) Sₙ = a(rⁿ − 1)/(r − 1) · S∞ = a/(1 − r), |r| < 1

23Special sums

In plain English: handy closed forms so you never add 1+2+…+n by hand; remember the cube-sum equals the square of the plain sum.
Sum of first n natural numbers, their squares and cubes.
Note: Σn³ = (Σn)².
e.g. 1+2+…+10 = 10×11/2 = 55; and 1³+2³+…+10³ = 55² = 3025.

Σn = n(n + 1)/2 Σn² = n(n + 1)(2n + 1)/6 Σn³ = [n(n + 1)/2]²

24Means & counting handshakes

In plain English: the three means rank AM ≥ GM ≥ HM; and any "everyone meets everyone once" count (handshakes, matches, lines) is just nC2 pairs.
AM-GM-HM for two positives a, b; AM ≥ GM ≥ HM.
Pairs from n people (handshakes / matches / lines / diagonals).
e.g. 6 people each shake hands once: ⁶C₂ = 15 handshakes. For a, b = 4, 9: AM = 6.5, GM = 6.

AM = (a+b)/2 · GM = √(ab) · HM = 2ab/(a+b) Pairs = nC2 = n(n − 1)/2 · Diagonals = n(n − 3)/2

Probability, CAT PYQs

Probability

Favourable-over-total counting cast as probability.

CAT 1995

EasyCAT 1995

If a 4 digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?

(1) 1/12
(2) 1/4
(3) 1/6
(4) None of these

Show solution

(1) 1/12. Total 4-digit numbers = 4! = 24. For divisibility by 25 the last two digits must be 25, so the first two digits can be arranged in 2! = 2 ways. Probability = 2!/4! = 2/24 = 1/12.