
Find how many numbers can be formed with the digits \[1,2,3,4,3,2,1\] so that the odd digits always occupy the odd places.
A. \[18\]
B. \[28\]
C. 6
D. \[27\]
Answer
164.4k+ views
Hint: First, find the number of odd and even numbers present in the given digits. Then, arrange the digits in even and odd places. In the end, find how many ways the digits can be arranged to get the required answer.
Formula Used: Permutation formula when repetition allowed: \[\dfrac{{n!}}{{{a_1}!{a_2}!....{a_n}!}}\]
Complete step by step solution: The given digits are \[1,2,3,4,3,2,1\].
Total number of digits: 7
Number of odd numbers: \[1,3,3,1\]: 4
Number of even numbers: \[2,4,2\]: 3
Since in the given digits 3 even numbers \[2,4,2\] are present. There are 3 even places.
And the digit 2 is repeated 2 times.
So, the number of ways of arranging the even digits at 3 places are: \[\dfrac{{3!}}{{2!}} = 3\]
Also, there are 4 even numbers \[1,3,3,1\] present and we must arrange them into 4 places.
But both numbers repeated 2 times.
So, the number of ways of arranging the odd digits at 4 places are: \[\dfrac{{4!}}{{2!2!}} = 6\]
Therefore, the number of words formed in which vowels occupy the even places are:
\[\dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2!2!}} = 3 \times 6\]
\[ \Rightarrow \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2!2!}} = 18\]
Option ‘C’ is correct
Note: Permutation shows the number of possible arrangements of the objects when the order of the arrangement of the objects matters.
If some objects are repeated, then apply the formula of the permutation for the repetition.
Formula Used: Permutation formula when repetition allowed: \[\dfrac{{n!}}{{{a_1}!{a_2}!....{a_n}!}}\]
Complete step by step solution: The given digits are \[1,2,3,4,3,2,1\].
Total number of digits: 7
Number of odd numbers: \[1,3,3,1\]: 4
Number of even numbers: \[2,4,2\]: 3
Since in the given digits 3 even numbers \[2,4,2\] are present. There are 3 even places.
And the digit 2 is repeated 2 times.
So, the number of ways of arranging the even digits at 3 places are: \[\dfrac{{3!}}{{2!}} = 3\]
Also, there are 4 even numbers \[1,3,3,1\] present and we must arrange them into 4 places.
But both numbers repeated 2 times.
So, the number of ways of arranging the odd digits at 4 places are: \[\dfrac{{4!}}{{2!2!}} = 6\]
Therefore, the number of words formed in which vowels occupy the even places are:
\[\dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2!2!}} = 3 \times 6\]
\[ \Rightarrow \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2!2!}} = 18\]
Option ‘C’ is correct
Note: Permutation shows the number of possible arrangements of the objects when the order of the arrangement of the objects matters.
If some objects are repeated, then apply the formula of the permutation for the repetition.
Recently Updated Pages
JEE Advanced Percentile vs Marks 2025| Previous year's trends

JEE Advanced 2021 Physics Question Paper 2 with Solutions

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2025 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced Marks vs Rank 2025 - Predict IIT Rank Based on Score

Trending doubts
JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Advanced Exam Pattern 2025

MHT CET 2025: Exam Date PCM and PCB (OUT), Application Form (Open), Eligibility and Syllabus Updates

IIT Kanpur Highest Package, Average & Median Salary

JEE Advanced 2025 Reservation Criteria: SC, ST, OBC, PwD, and EWS

Other Pages
Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
