Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Permutations and Combinations question

2024 · 29 Jan · Shift 1 · Q57
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Permutations and Combinations
  5. /2024 · 29 Jan · Shift 1 · Q57

Permutations and Combinations question

2024 · 29 Jan · Shift 1 · Q57

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 553

  1. Arrange the letters in dictionary order

The word is GTWENTY.

Its letters are: G,T,W,E,N,T,YG, T, W, E, N, T, YG,T,W,E,N,T,Y

In alphabetical order, these are: E<G<N<T<T<W<YE < G < N < T < T < W < YE<G<N<T<T<W<Y

We need the rank (serial number) of the arrangement: GTWENTYGTWENTYGTWENTY among all distinct permutations of these letters, arranged lexicographically.

Since there are two identical TTT's, we must divide by 2!2!2! whenever all remaining letters include both TTT's.


  1. Count words before GTWENTYGTWENTYGTWENTY letter by letter

We count how many valid distinct arrangements come before it.

Position 1: GGG

Available letters: E,G,N,T,T,W,YE, G, N, T, T, W, YE,G,N,T,T,W,Y

Letters smaller than GGG are only EEE.

If first letter is EEE, remaining letters are G,N,T,T,W,YG,N,T,T,W,YG,N,T,T,W,Y. Number of arrangements: 6!2!=7202=360\frac{6!}{2!}=\frac{720}{2}=3602!6!​=2720​=360

So words before ours from position 1: 360360360


Position 2: TTT

Now first letter fixed as GGG. Remaining letters: E,N,T,T,W,YE,N,T,T,W,YE,N,T,T,W,Y

We need letters smaller than TTT: E,NE, NE,N.

  • If second letter is EEE, remaining letters: N,T,T,W,YN,T,T,W,YN,T,T,W,Y 5!2!=60\frac{5!}{2!}=602!5!​=60
  • If second letter is NNN, remaining letters: E,T,T,W,YE,T,T,W,YE,T,T,W,Y 5!2!=60\frac{5!}{2!}=602!5!​=60

Total before ours from position 2: 60+60=12060+60=12060+60=120

Running total: 360+120=480360+120=480360+120=480


Position 3: WWW

Prefix fixed: GTGTGT Remaining letters: E,N,T,W,YE,N,T,W,YE,N,T,W,Y

Letters smaller than WWW: E,N,TE,N,TE,N,T.

Each choice leaves 4 distinct letters, so each contributes: 4!=244!=244!=24

Thus total contribution: 3×24=723\times 24=723×24=72

Running total: 480+72=552480+72=552480+72=552


Position 4: EEE

Prefix fixed: GTWGTWGTW Remaining letters: E,N,T,YE,N,T,YE,N,T,Y

There is no letter smaller than EEE.

Contribution: 000


Position 5: NNN

Prefix fixed: GTWEGTWEGTWE Remaining letters: N,T,YN,T,YN,T,Y

No letter smaller than NNN among remaining.

Contribution: 000


Position 6: TTT

Prefix fixed: GTWENGTWENGTWEN Remaining letters: T,YT,YT,Y

No letter smaller than TTT among remaining.

Contribution: 000


Position 7: YYY

Only one letter remains, so contribution is 000.


  1. Find the serial number

Number of words before GTWENTYGTWENTYGTWENTY is: 552552552

Therefore its rank is: 552+1=553552+1=553552+1=553


  1. Comparison with stored correct answer

Stored correct answer = 553553553

Our derived answer is also: 553553553

So the stored answer is correct.

PreviousNext

More from Permutations and Combinations

  • Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to2024 · MCQ
  • In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : A,B and C. A student is required to attempt total 15 questions taking at least 4 questions from each…2024 · Numerical
  • The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to ​.2024 · Numerical
  • The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is2024 · MCQ
  • If for some m,n;6Cm​+2(6Cm+1​)+6Cm+2​>8C3​ and n−1P3​:nP4​=1:8, then nPm+1​+n+1Cm​ is equal to2024 · MCQ
  • The value of 1!50!1​+3!48!1​+5!46!1​+….+49!2!1​+51!1!1​ is :2023 · MCQ
  • The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7, is ​.2023 · Numerical
  • The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ​.2023 · Numerical