How many numbers can be formed with the digits 1 2 3 4 3 2 1 Taken all together so that the odd digits always occupy the odd places?

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?

Solution

There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in\[\frac{4!}{2!2!}\]ways.
The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in\[\frac{3!}{2!}\]ways.
By fundamental principle of counting:
Required number of arrangements =\[\frac{4!}{2!2!}\]\[\times\]\[\frac{3!}{2!}\]= 18

Concept: Factorial N (N!) Permutations and Combinations

  Is there an error in this question or solution?

APPEARS IN

Solution : We have been given seven digits, namely 1, 2, 3, 4, 3, 2, 1.
So, we have to form 7-digit numbers, so that odd digits occupy odd places.
Every 7 - digits number has 4 odd places.
Given four odd digits are 1, 3, 3, 1 out of which 1 occurs 2 times and 3 occurs 2 times.
So, the number of ways to fill up 4 odd places `=(4!)/((2!)xx(2!))=6.`
Given three even digits are 2, 4, 2 in which 2 occurs 2 times and 4 occurs 1 time.
So, the number of ways to fill up 3 even places`=(3!)/(2!)=3.`
Hence, the required number of numbers `=(6xx3)= 18.`

How many numbers of 3 digits can be formed with the digits 1 2 3 4 5 repetition of digits are not allowed?

How many numbers can be formed by using all the digits 1 2 3 4?

How many 3

How many 3