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

2020 · 7 Jan · Shift 1 · Q23
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. /2020 · 7 Jan · Shift 1 · Q23

Permutations and Combinations question

2020 · 7 Jan · Shift 1 · Q23

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
  1. A
    52(6!){5 \over 2}\left( {6!} \right)25​(6!)
  2. B
    6!{6!}6!
  3. C
    56
  4. D
    12(6!){1 \over 2}\left( {6!} \right)21​(6!)
View written solutionFree

Correct answer: A

  1. We need the number of 6-digit numbers formed using only the digits 1,3,5,7,91,3,5,7,91,3,5,7,9 and such that all five digits appear.

  2. Since there are 6 positions but only 5 distinct digits available, and all five must appear, exactly one digit must be repeated once.

  3. Choose the digit that is repeated: 5 ways5 \text{ ways}5 ways

  4. Now arrange these 6 symbols, where one digit is repeated twice and the other four digits appear once each.

The number of distinct arrangements is: 6!2!\frac{6!}{2!}2!6!​

  1. Therefore total numbers: 5⋅6!2!=52(6!)5 \cdot \frac{6!}{2!} = \frac{5}{2}(6!)5⋅2!6!​=25​(6!)

  2. Compute if desired: 52⋅720=5⋅360=1800\frac{5}{2}\cdot 720 = 5\cdot 360 = 180025​⋅720=5⋅360=1800

So the required number is: 52(6!)\frac{5}{2}(6!)25​(6!)

  1. Checking options:
  • A: 52(6!)\dfrac{5}{2}(6!)25​(6!) ✅
  • B: 6!6!6! ❌
  • C: 565656 ❌
  • D: 12(6!)\dfrac{1}{2}(6!)21​(6!) ❌

Hence, the correct option is A.

PreviousNext

More from Permutations and Combinations

  • The number of ordered pairs (r, k) for which 6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :2020 · MCQ
  • An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ​.2020 · Numerical
  • If a, b and c are the greatest value of 19Cp, 20Cq and 21Cr respectively, then :2020 · MCQ
  • The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word 'EXAMINATION' is ​.2020 · Numerical
  • If the number of five digit numbers with distinct digits and 2 at the 10th place is 336 k, then k is equal to :2020 · MCQ
  • All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is :2019 · MCQ
  • The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0,1,2,3,4,5 (repetition of digits is allowed) is :2019 · MCQ
  • A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then :2019 · MCQ