How many number greater than 10 lac be formed from \(2,3,0,3,4,2,3\) (a) 420 (b) 360 (c) 400 (d) 300

Short Answer

Expert verified
The short answer to the question is 400, which corresponds to option (c).

Step by step solution

01

Identify the total possible permutations

Since we have 7 digits, we can create 7! permutations of them. However, we have duplicate digits, which we must account for in our calculation. We have two 2's and three 3's, so we need to divide by (2! * 3!) to get the total possible permutations: \[\frac{7!}{(2! * 3!)}\]
02

Calculate the total permutations

Now, calculate the total possible permutations using the formula above: \[ \frac{7!}{(2! * 3!)} = \frac{7*6*5*4*3*2*1}{(2*1)*(3*2*1)} = \frac{5040}{(2)*(6)} = 420\]
03

Determine how many permutations are greater than 1 million

From the total permutations obtained in step 2, not all of them are greater than 1 million (10 lac). If the first digit is 0 or 1, the number will be smaller than 1 million. Since we cannot change the order of the digits, the only way the number is greater than 1 million is when it starts with 2, 3, or 4. However, because we cannot change the order of the digits, 2 and 3 must appear before 4; thus, the number will always start with 2 or 3. - If the number starts with 2, we have 6 remaining digits: 3, 0, 3, 4, 3, 3. Using the same permutation principles as in step 1, we get: \[\frac{6!}{(3!)} = \frac{6*5*4*3*2*1}{(3*2*1)} = 120\] - If the number starts with 3, we have 6 remaining digits: 2, 0, 3, 4, 2, 3. Using the permutation formula we get: \[\frac{6!}{(2!)} = \frac{6*5*4*3*2*1}{(2*1)} = 360\]
04

Find the total number of numbers meeting the criteria

To find the total number of 7-digit numbers greater than 1 million, add the two values obtained in step 3: 120 + 360 = 480. However, in both cases, we have included the numbers where 0 appears as the second digit (which is not valid for a 7-digit number). We need to subtract such cases from our previous answer to get the final result. - If the number starts with 2 and the second digit is 0, we have 5 remaining digits: 3, 3, 4, 3, 3, and calculate the permutations: \[\frac{5!}{(3!)} = \frac{5*4*3*2*1}{(3*2*1)} = 20\] - If the number starts with 3 and the second digit is 0, we have 5 remaining digits: 2, 3, 4, 2, 3 and calculate the permutations: \[\frac{5!}{(2!)} = \frac{5*4*3*2*1}{(2*1)} = 60\] Subtract both cases from the previous answer and we're left with: 480 - (20 + 60) = 400.
05

Answer

The correct answer is 400. So, the student should choose option (c).

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