Given:
The digits 1, 2, 3, 4, 3, 2, 1
The total number of digits are 7.
There are 4 odd digits 1,1,3,3 and 4 odd places (1,3,5,7)
So, the odd digits can be arranged in odd places in = \( \dfrac{4!}{(2! 2!)}\) ways.
The remaining even digits 2,2,4 can be arranged in 3 even places in = \(\dfrac{ 3!}{2!}\) Ways.
Hence, the total number of digits =\( \dfrac{4!}{(2! 2!)}\) × \( \dfrac{ 3!}{2!}\)
= 3×2×1×3×1
= 18
Hence, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.
Answered by Sakshi | 1 year agoHow many words each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
How many words, with or without meaning can be formed from the letters of the word ‘MONDAY’, assuming that no letter is repeated, if
(i) 4 letters are used at a time
(ii) all letters are used at a time
(iii) all letters are used but first letter is a vowel ?
There are 10 persons named P1, P2, P3 …, P10. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.
How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?