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The number of different numbers that can...

The number of different numbers that can be formed by using all the digits 1, 2, 3, 4,3, 2,1 so that odd digits always occupy the odd places IS 1) 6 2)72 3) 60 4) 18

A

3!4!

B

34

C

18

D

12

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the number of different numbers that can be formed using the digits 1, 2, 3, 4, 3, 2, 1, such that odd digits always occupy the odd places, we can follow these steps: ### Step 1: Identify the digits and their positions We have the digits: 1, 2, 3, 4, 3, 2, 1. The total number of digits is 7. The odd positions in a 7-digit number are the 1st, 3rd, 5th, and 7th positions. ### Step 2: Identify the odd and even digits The odd digits from the set are: 1, 1, 3, 3 (total of 4 odd digits). The even digits from the set are: 2, 2, 4 (total of 3 even digits). ### Step 3: Arrange the odd digits in the odd positions We have 4 odd positions and 4 odd digits. The number of ways to arrange these odd digits is calculated using the formula for permutations of multiset: \[ \text{Ways to arrange odd digits} = \frac{4!}{2! \times 2!} \] Here, \(4!\) accounts for the total arrangements of the 4 digits, and we divide by \(2!\) for the two indistinguishable 1s and \(2!\) for the two indistinguishable 3s. Calculating this gives: \[ 4! = 24 \] \[ 2! = 2 \] \[ \text{Ways to arrange odd digits} = \frac{24}{2 \times 2} = \frac{24}{4} = 6 \] ### Step 4: Arrange the even digits in the even positions Next, we have 3 even positions and 3 even digits (2, 2, 4). The number of ways to arrange these even digits is: \[ \text{Ways to arrange even digits} = \frac{3!}{2!} \] Calculating this gives: \[ 3! = 6 \] \[ \text{Ways to arrange even digits} = \frac{6}{2} = 3 \] ### Step 5: Calculate the total arrangements The total number of different numbers that can be formed is the product of the arrangements of odd digits and even digits: \[ \text{Total arrangements} = \text{Ways to arrange odd digits} \times \text{Ways to arrange even digits} = 6 \times 3 = 18 \] ### Final Answer Thus, the number of different numbers that can be formed is **18**. ---
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