Question

How many natural numbers not exceeding
4321 can be formed with the digits 1, 2, 3.4
if the repitition of digits in a number is
not allowed ?


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Answer 1

Answer:

313

Step-by-step explanation:

Anyway, there are 4 one digit numbers (obvious)

16 two digit numbers (there are 4 options for each digit, and two digit numbers are automatically <= 4321)

64 three digit numbers (again, 4 options for each digit)

Now for the four digit numbers:

There are 64 numbers where the first digit is a 3 (4 options for each of the remaining digits, and since the first digit is a 3, the number is < 4321)

Similarly there are 64 numbers where the first digit is a 2, and another 64 where the first digit is a 1.

This is 192 four digit numbers so long.

Now lets look at what happens if the first digit is a 4.

There are 16 numbers where the second digit is a 2 (the last two digits can be anything in this case) and another 16 where the second digit is a 1.

If the second digit is a 3:

...If the third digit is a 1, there are 4 numbers, because there are no restrictions on the last digit.

...If the third digit is a 2, there is only one possible number, n.l. 4321.

So, there are 16 numbers if the second digit is a 2, 16 numbers if the second digit is a 1, and 4 + 1 = 5 numbers if the second digit is a 3, giving

16 + 16 + 5 = 37 four digit numbers where the first digit is a 4.

In total, we have

4 one digit numbers

16 two digit numbers

64 three digit numbers

192 four digit numbers where the first digit is not a 4

37 four digit numbers where the first digit is a 4

This is 313 numbers in total.

Answer 2

Answer:

313 numbers

Step-by-step explanation: