Question

The number of numbers less than 1000 that can be formed out of the digits 0 1 2 3 4 5 no digit being repeated

Answer 1

We have to find out how many one-digit, two-digit and three-digit integers can be formed using the numbers 0, 1, 2, 3, 4 and 5 without digit repetition.

⇒  0 can't be taken as a one-digit integer. So 1, 2, 3, 4 and 5 can only be taken. Thus the no. of one-digit integers is 5.

⇒  In the places of a two-digit integer,

    →  0 can't be taken at the tens place. Thus the other 5 numbers can only be taken.

    →  If a number is used at tens place, the other 5 numbers including 0 can be taken at ones place.

    Hence the no. of two-digit integers is 5 × 5 = 25.

⇒  In the places of a three-digit integer,

    →  0 can't be taken at the hundreds place. Thus the other 5 numbers can only be taken.

    →  If a number is used at hundreds place, the other 5 numbers including 0 can be taken at tens place.

    →  If two numbers are used at hundreds and tens place each, the other 4 numbers (sometimes 0 includes) can be taken at tens place.

    Hence the no. of three-digit integers is 5 × 5 × 4 = 100.

Hence the total no. of integers we can form is 5 + 25 + 100 = 130.

Here's the method according to permutation:

No. of one-digit integers = ⁵P₁.

No. of two digit integers = ⁵P₁ × ⁵P₁.

No. of three-digit integers = ⁵P₁ × ⁵P₂.

Total no. of integers

= ⁵P₁ + ⁵P₁ × ⁵P₁ + ⁵P₁ × ⁵P₂

= ⁵P₁ (1 + ⁵P₁ + ⁵P₂)

= 5 (1 + 5 + 20)

= 5 × 26

= 130

Answer 2

$\huge{\underline{\underline{\textbf{Solution:-}}}}$






The Numbers are $\sf 0,1, 2, 3, 4, 5$

1. 3 digit numbers:

Choose the first digit any of 5 ways. (can't be 0)

Choose the second digit any of 5 ways. (can be 0)

Choose the second digit any of 4 ways.

That's (5)(5)(4) = 100 3 digit numbers

2. 2 digit numbers:

Choose the first digit any of 5 ways. (can't be 0)

Choose the second digit any of 5 ways. (can be 0)

That's (5)(5) = 25 2 digit numbers

3. 1-digit numbers:

6 ways 0,1,2,3,4,5

Total: 100 + 25 + 6 = 131