# Understand the problem

(a)2439. (b)4096. (c)4903. (d)4904. (e)5416

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Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

**Step 1.** After having a look into this problem you can see our main aim here is to find out the total number of 4 digit positive numbers with at least one 2 or one 3 in it. Now we can do this by subtracting the number of 4 digit positive number which do not have any 2’s or 3’s from the total number of 4 digit positive numbers. Its easy, give it a try.

**Step 2.** Now lets’ proceed forward and try to find out the total number of 4 digit positive integers . We can use all digit from 0 to 9 as we do not have any restriction . Now to make a table with this

**Step 3.** Now for the 2nd part. So try to find out the total number of 4 digit positive integers which don’t have any 2’s and 3’s in it . So now in this case we have 8 digits(2 and 3 are excluded). Now to make a table with this

**Step 4.** Now when you have found the number of positive 4 digit numbers in both the cases now

(total number of 4 digit positive integers)-( total number of 4 digit positive integers which don’t have any 2’s and 3’s in it )=(total number of 4 digit positive numbers with at least one 2 or one 3 in it). Very close to the solution just about to crack it !!!.

Total number of 4 digit positive integers(Table 1) = 9000 Total number of 4 digit positive integers which don’t have any 2’s and 3’s in it(Table 2) = 3584 Total number of 4 digit positive numbers with at least one 2 or one 3 in it = 5416 that’s your required answer.

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