Question

The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is

पर्याय

• 127

• 128

• 502

• 501

Answer 1

502

Explanation:

Note the following constraint:

None of the digits in the following cases can be 0.

The reason:

Since the digits must increase from left to right, any combination that includes 0 must place 0 in the leftmost position --not possible, since the leftmost digit of a positive integer must be at least 1.

Thus, all of the following cases must be formed from the following pool of digits: 1, 2, 3, 4, 5, 6, 7, 8, 9.

We should also note the following:

ⁿC_r = ⁿC_(n–r)

To illustrate:

⁶C₅= ⁶C_{(6 –5)}= ⁶C₁.

From 6 options, the number of ways to choose 5 = ⁶C₅= (6→5→4→3→2)/(5→4→3→2→1) = 6.

From 6 options, the number of ways to choose 1 = ⁶C₁= 6.

Thus:

⁶C₅= ⁶C₁.

2-digit integers:

Number of ways to choose 2 distinct integers from 9 options = ⁹C₂= (9→8)/(2→1) = 36.

3-digit integers:

Number of ways to choose 3 distinct integers from 9 options = ⁹C₃= (9→8→7)/(3→2→1) = 84.

4-digit integers:

Number of ways to choose 4 distinct integers from 9 options = ⁹C₄= (9→8→7→6)/(4→3→2→1) = 126.

5-digit integers:

Number of ways to choose 5 distinct integers from 9 options = ⁹C₅= ⁹C₄= 126.

6-digit integers:

Number of ways tochoose 6 distinct integers from 9 options = ⁹C₆= ⁹C₃= 84.

7-digit integers:

Number of ways to choose 7 distinct integers from 9 options = ⁹C₇= ⁹C₂= 36.

8-digit integers:

Number of ways to choose 8 distinct integers from 9 options = ⁹C₈ = ⁹C₁= 9.

9-digit integers:

Here, there is only 1 option:

123456789.

 

Total options = 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1 = 502