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**Hint:**We divide the 30 numbers in three types of numbers using the divisibility by 3. Then calculate the number of ways of selecting two numbers from all 30 by taking two cases one by one. First choose both numbers from the set where all numbers are divisible by 3, then the second case where one number from one set and another is from the second set. Use the method of combinations to calculate the number of ways of selecting two numbers.

* Combination formula is given by\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where n is total number of items w are choosing from and r is number of items we are selecting.

* Factorial terms open up as \[n! = n(n - 1)!\]

**Complete answer:**

We are given the set of numbers \[\left\{ {1,2,....30} \right\}\]

We divide the numbers from the set into three forms i.e. \[3k,3k + 1,3k + 2\], where k is any positive integer.

Each number in the set falls in one of the categories.

We write numbers falling in each category separately.

\[3k:3,6,9,12,15,18,21,24,27,30\]

\[3k + 1:1,4,7,10,13,16,19,22,25,28\]

\[3k + 2:2,5,8,11,14,17,20,23,26,29\]

There are 10 numbers falling in each category.

Now we make cases for selecting numbers from categories.

Case1: Choosing both the numbers from 3k category

If we take both the numbers from 3k category then their sum will be divisible by 3 as \[3{k_1} + 3{k_2} = 3({k_1} + {k_2})\]

So we found a number of ways to select both numbers from category 3k.

Here n is 10 and r is 2

\[ \Rightarrow \]Number of ways of selecting two numbers \[{ = ^{10}}{C_2}\]

Use the combinations method to find the value

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = \dfrac{{10!}}{{(10 - 2)!2!}}\]

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = \dfrac{{10!}}{{8!2!}}\]

Open the term in numerator using factorial method

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = \dfrac{{10 \times 9 \times 8!}}{{8!2!}}\]

Cancel same terms from numerator and denominator

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = 45\] … (1)

Case2: Choosing one number from \[3k + 1\] and another from \[3k + 2\]

If we one number from \[3k + 1\]and another from\[3k + 2\], then their sum will be divisible by 3 as

\[3{k_1} + 1 + 3{k_2} + 2 = 3{k_1} + 3{k_2} + 3 = 3({k_1} + {k_2} + 1)\]

So we find number of ways to select one number from \[3k + 1\]and another from \[3k + 2\]

Here n is 10 and r is 1 for each case

\[ \Rightarrow \]Number of ways of selecting two numbers \[{ = ^{10}}{C_1}{ \times ^{10}}{C_1}\]

Use the combinations method to find the value

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = \dfrac{{10!}}{{(10 - 1)!1!}} \times \dfrac{{10!}}{{(10 - 1)!1!}}\]

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = \dfrac{{10!}}{{9!1!}} \times \dfrac{{10!}}{{9!1!}}\]

Open the term in numerator using factorial method

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = \dfrac{{10 \times 9!}}{{9!}} \times \dfrac{{10 \times 9!}}{{9!}}\]

Cancel same terms from numerator and denominator

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = 100\] … (2)

Total number of ways can be calculated by taking the sum from both the cases.

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = 45 + 100\]

\[ \Rightarrow \]Number of ways of selecting two numbers \[ = 145\]

**\[\therefore \]Correct option is B.**

**Note:**

Many students try to solve this question by taking each and every case of two numbers which is very lengthy as there are 30 initial numbers and we have to take possibility for each pair and then check if it is divisible by 3 or not. Kindly use the approach of combination for such questions.

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