Permutations and Combinations

Maths · Class 11

Simulation · Maths · Class 11

Tag the twins, then take the tags off

"Tag the twins, then take the tags off" is a free, interactive Class 11 Maths simulation for JEE, built on the NCERT section "Permutations with repeated objects" in Permutations and Combinations. Arrangements of a word's letters with subscripts on repeated letters (L₁, L₂); removing the labels merges groups of p! identical-looking arrangements into one. JEE Main asked 47 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).

From the lesson Permutations with repeated objects in Permutations and Combinations. Change the values and watch what happens.

Tag the twins, then take the tags offMaths · Class 11

What it shows

NCERT §6.3.4

Arrangements of a word's letters with subscripts on repeated letters (L₁, L₂); removing the labels merges groups of p! identical-looking arrangements into one.

Controls

3 to try

  • Segmentedword BALL / LEVEL / BALLOON (sample)
  • Togglelabels on / off
  • Tapone merged arrangement to see its group

Try this

With BALLOON, turn labels off and read the count 1260 = 7!/(2! × 2!).

Common mistake

Dividing only by the count of one repeated letter.

In JEE Main

42 shifts analysed

47 questions from Permutations and Combinations across the 42 JEE Main shifts Lumi has analysed (2017–2026).

See the past questions

The idea behind it

NCERT §6.3.4

  • If some objects are alike, swapping them does not give a new arrangement, so dividing by the number of such swaps removes the repeats.
  • The number of arrangements of n objects of which p₁ are alike of one kind, p₂ alike of another kind, and so on, is n!/(p₁! p₂! … pₖ!).
  • BALLOON has 7 letters with L twice and O twice, so it has 7!/(2! × 2!) = 5040/4 = 1260 arrangements.
  • Pennants: 3 red, 2 blue and 1 green pennant in a row, same colours alike, give 6!/(3! × 2! × 1!) = 720/12 = 60 patterns.
  • SPORTS has S twice: 6!/2! = 360 arrangements. Arrangements of BALLOON that start with B: fix B and arrange A, L, L, O, O, N, giving 6!/(2! × 2!) = 180.
  • Check with labels: tag the two Ls as L₁ and L₂. Each arrangement with plain Ls matches 2! labelled ones, which is why we divide by 2!.

Questions

4 answers

What does the "Tag the twins, then take the tags off" simulation show?
Arrangements of a word's letters with subscripts on repeated letters (L₁, L₂); removing the labels merges groups of p! identical-looking arrangements into one.
What mistake does the "Tag the twins, then take the tags off" simulation fix?
A common mistake it clears up: dividing only by the count of one repeated letter.
Does JEE Main ask questions from Permutations and Combinations?
Yes. JEE Main asked 47 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
Is this simulation free?
Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Permutations with repeated objects" in Permutations and Combinations (Class 11 Maths).

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