How many numbers can be formed with the digits 1,2,3,4,3,2 and 1 so that the odd digits always occupy odd places? Answer is 18. I have considered from 2 digit number to 7 digit number since there is no point in talking about 1 digit number.Also, 8 digit number does not make any sense since there there are only 3 even numbers and 4 even places.Like this I am getting answer more than 18.Where am I wrong?
It is clear from statement that we have to use all the given digits,so numbers must be of 7 digits.Since,odd digits occupy odd places,so, in a 7 digits number let us fix 1st, 3rd, 5th, & 7th place for odd digit nos. and given odd nos. are 1,3,3,1.So, they can arrange in 4 places mentioned above in 4!/2!.2! = 6 ways.Similiarly even nos. can arrange in 3 ways.So, total numbers obt. are 6X3=18