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Rzqust [24]
3 years ago
11

Find the number of positive integers less than 100,000 whose digits are among 1, 2, 3, and 4.

Mathematics
2 answers:
Katen [24]3 years ago
7 0

Answer:

1364

Step-by-step explanation:

Paraphin [41]3 years ago
4 0

As we can see that there are 6 digits in 100,000 and its is the smallest number we can have in 6 digit. So all numbers less than 100,000 will be 1-digit, 2-digits, 3-digits, 4-digits and 5-digits numbers made from 1,2,3,4 with repetitions allowed.

Case 1: All 1 -digit numbers

We will have numbers 1,2,3,4. So total 4 integers for this one

Case2: All 2-digit numbers

We can fill 1 digit place in 4 ways ( can choose any number out of 1,2,3,4). Then again we can fill 2nd digit place in 4 ways ( can choose any number out of 1,2,3,4). So all together we will have 4 × 4 = 16 integers for this one

Case3: All 3-digits numbers

We can fill 1 digit place in 4 ways ( can choose any number out of 1,2,3,4). Then again we can fill 2nd digit place in 4 ways ( can choose any number out of 1,2,3,4). similarly we can fill 3rd digit place in 4 ways (again any number out of 1,2,3,4). So all together we will have 4 × 4 × 4 = 64 integers for this one.

Case4: All 4-digit numbers

Again we can fill 1st digit place in 4 ways, then 2nd digit place in 4 ways, 3rd digit place in 4 ways, 4th digit place in 4 ways. So all together there will be 4 × 4 × 4 × 4 = 256 integers for this one.

Case5: All 5-digit numbers

Again we can fill 1st digit place in 4 ways, then 2nd digit place in 4 ways, 3rd digit place in 4 ways, 4th digit place in 4 ways, 5th digit place also in 4 ways. So all together there will be 4 × 4 × 4 × 4 × 4 = 1024 integers for this one

Adding results of all 5 cases we get,

Total integers = 4 + 16 +64 + 256 + 1024 = 1364 integers.

So thats the final answer

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<h3>How to solve the inequality?</h3>

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