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JEE Mains Maths · Permutations and Combinations

Selections and Committees

Choosing groups when order does not matter — committees with 'at least' conditions, picks from several groups, letters chosen from a word and then arranged — and simplifying expressions in nCr and nPr.

All Permutations and Combinations notesJEE Mains Maths strategy

Why this matters

Twenty-five PYQs, fourteen of them numerical answer. Most split into cases by how many come from each group and add products of combinations. Six choose letters from a word with repeats, and five are algebra with the nCr and nPr formulas. Three ideas cover the page.

Concept 1 of 3: Committees and 'at least' conditions

List the allowed compositions — how many from each group — and for each multiply the combinations, then add. 'At least one' is often faster as total minus none. When a person must be included, remove them and choose the rest; when two may not both be in, subtract the selections containing both.

Definition

  • Compositions (a,b)(a,b) from groups of sizes m,nm,n: ∑(ma)(nb)\sum\binom ma\binom nb.
  • At least one of a kind == total −- none of that kind.
  • A fixed person included: choose the remaining r−1r-1 from n−1n-1.
  • Two people not both in: (nr)−(n−2r−2)\binom nr-\binom{n-2}{r-2}.

Case sum

∑allowed (a,b)(ma)(nb)\sum_{\text{allowed }(a,b)}\binom ma\binom nb

Worked example

A team of 4 is chosen from 5 men and 4 women with at least 2 women. In how many ways?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 1 · Q62Moderate

Example 1 · Permutations and Combinations · Selections and Committees

A scientific committee is to formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:

Do not choose 'at least one' first

Choosing one woman first and then any others counts the same committee several times. Split into exact cases, or use the complement.
Drill 13 more on committees and 'at least' conditions

Concept 2 of 3: Choosing letters from a word

When letters are taken from a word with repeats, split by pattern: all different, one pair and the rest different, two pairs, a triple, and so on. For each pattern count the choices of letters, then the arrangements with that pattern. For selections only (not words), stop before arranging.

Definition

  • Patterns for 4 letters: all different, 2+1+12+1+1, 2+22+2, 3+13+1, 44.
  • Arrangements of a pattern: 4!2!\frac{4!}{2!}, 4!2! 2!\frac{4!}{2!\,2!}, 4!3!\frac{4!}{3!}.
  • Vowels and consonants chosen separately: multiply, then arrange.

One pattern

(choices of letters)×r!(repeats)!\text{(choices of letters)}\times\frac{r!}{\text{(repeats)}!}

Worked example

How many 3-letter words can be made from the letters of APPLE?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q175Moderate

Example 2 · Permutations and Combinations · Selections and Committees

The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is

Distinct letters, not letter count

UNIVERSE has 8 letters but E repeats; without repetition there are only 3 distinct vowels. Count distinct letters before choosing.
Drill 5 more on choosing letters from a word

Concept 3 of 3: Working with the nCr and nPr formulas

Ratios of permutation or combination symbols cancel to short products: nPrn−1Pr−1=n\frac{{}^nP_r}{{}^{n-1}P_{r-1}}=n, 2nC3nC3=4(2n−1)n−2\frac{{}^{2n}C_3}{{}^nC_3}=\frac{4(2n-1)}{n-2}. Pascal's rule combines neighbours. A sum like ∑k⋅k!\sum k\cdot k! telescopes because k⋅k!=(k+1)!−k!k\cdot k!=(k+1)!-k!.

Definition

  • nPr=n!(n−r)!{}^nP_r=\frac{n!}{(n-r)!}, nCr=n!r! (n−r)!{}^nC_r=\frac{n!}{r!\,(n-r)!}.
  • (nr)+2(nr+1)+(nr+2)=(n+2r+2)\binom nr+2\binom n{r+1}+\binom n{r+2}=\binom{n+2}{r+2}.
  • k⋅k!=(k+1)!−k!k\cdot k!=(k+1)!-k!.
  • (mn)!(m!)n\frac{(mn)!}{(m!)^{n}} is an integer (it counts arrangements into labelled groups).

Telescoping

∑k=1nk⋅k!=(n+1)!−1\sum_{k=1}^{n}k\cdot k!=(n+1)!-1

Worked example

If nC2=45{}^nC_2=45, find nn.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q63Moderate

Example 3 · Permutations and Combinations · Selections and Committees

If  2nC3: nC3=10:1\ ^{2n}C_{3}:\ ^{n}C_{3}= 10:1, then the ratio (n2+3n):(n2−3n+4)\left( n^{2}+ 3n \right):\left( n^{2}- 3n + 4 \right) is

Check which root is valid

Equations in nn from these formulas are often quadratic. Discard a root that is negative, not an integer, or smaller than rr.
Drill 4 more on working with the ncr and npr formulas

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Watch out for (3)

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