<p>How many numbers of seven digits can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places?</p>
Step-by-Step Solution
Key Concept: Identify that odd digits (1,1,3,3) must occupy odd positions (1st, 3rd, 5th, 7th) and even digits (2,2,4) must occupy even positions (2nd, 4th, 6th), then count arrangements with repetition using the multinomial coefficient.
<p><strong>Step 1: Identify odd and even digits.</strong></p><p>Odd digits: 1, 1, 3, 3 (four digits, two types with repetitions)</p><p>Even digits: 2, 2, 4 (three digits, two types with repetitions)</p><p><strong>Step 2: Identify odd and even positions.</strong></p><p>Odd positions (1st, 3rd, 5th, 7th): 4 positions</p><p>Even positions (2nd, 4th, 6th): 3 positions</p><p><strong>Step 3: Place odd digits in odd positions.</strong></p><p>Arrange {1, 1, 3, 3} in 4 odd positions = 4!/(2!×2!) = 24/4 = 6 ways</p><p><strong>Step 4: Place even digits in even positions.</strong></p><p>Arrange {2, 2, 4} in 3 even positions = 3!/2! = 6/2 = 3 ways</p><p><strong>Step 5: Apply multiplication principle.</strong></p><p>Total numbers = 6 × 3 = 18</p><p><strong>∴ Answer: 18</strong></p>
Correct Answer: 18