Question

The total number of ways of selecting two number from the set {1, 2, 3, 4, …., 3n} so that their sum is divisible by 3 is equal to
(a) \[\dfrac{2{{n}^{2}}-n}{2}\]
(b) \[\dfrac{3{{n}^{2}}-n}{2}\]
(c) \[2{{n}^{2}}-n\]
(d) \[3{{n}^{2}}-n\]

Answer 1

Hint: To solve this question, we should have some knowledge of the concept of the combination that is, \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] is a formula for choosing r elements out of n. Also, we need to know that every natural number can be expressed as either 3n or 3n – 1 or 3n – 2. By using them we can solve the question.

Complete step-by-step answer:
In this question, we have been asked to find the number of ways of solving two numbers from {1, 2, 3, …. 3n}, so that their sum is divisible by 3. To solve this question, we should know that if a number is divisible by 3 that means the number is a multiple of 3. So, if we say that sum should be divisible by 3 then the sum should be multiple of 3.
Now, we know that any natural number can be expressed as 3A or 3A – 1 or 3A – 2. So, from this set {1, 2, 3, …. 3n}
1, 4, 7, 10, ….., 3n – 2 is of 3A – 2 types and 2, 5, 8, 11, ….., 3n -1 is of 3A – 1 type and 3, 6, 9, 12, …. 3n is of 3A type.
Now, if we have to select two number whose sum to be divisible by 3 then we have 2 possible cases that are either both the numbers should be of 3A type or one number of (3A – 1) type and another of (3A – 2) type, because

Case 1: Both numbers are of 3A type. So, we can write 3m + 3m’ = 3 (m + m’) which is multiple of 3, hence divisible by 3.

Case 2: One number is of (3A – 1) type and another of (3A – 2) type. So, we can write 3m – 1 + 3m = 3(m + m’ – 1) which is again a multiple of 3, hence divisible by 3.

Now, we can say from the set {1, 2, 3, …. 3n} each type of number whether 3A or 3A – 1 or 3A – 2, there are n terms. So, if we have to choose numbers, then we will use the formula, \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] for choosing r elements from n elements for n = n in each type case.
According to case I, we have to choose the numbers of 3A type. So, the number of possible ways of choosing will be given for n = n and r = 2, that is \[^{n}{{C}_{2}}\].
And according to case II, we have to choose 1 number of (3A – 1) type and a number of (3A – 2) type. So, the possible ways of choosing can be given by \[\left( ^{n}{{C}_{1}} \right)\left( ^{n}{{C}_{1}} \right)\].

Hence, the total number of ways can be given by \[^{n}{{C}_{2}}+\left( ^{n}{{C}_{1}} \right)\left( ^{n}{{C}_{1}} \right)\] which can be expressed by using the formula \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. So, we get,
\[^{n}{{C}_{2}}+\left( ^{n}{{C}_{1}} \right)\left( ^{n}{{C}_{1}} \right)\]
\[\Rightarrow \dfrac{n!}{2!\left( n-2 \right)!}+\dfrac{n!}{\left( 1 \right)!\left( n-1 \right)!}\times \dfrac{n!}{1!\left( n-1 \right)!}\]
which can be further written as
\[\dfrac{n\left( n-1 \right)}{2}+n\times n\]
\[\dfrac{{{n}^{2}}-n}{2}+{{n}^{2}}\]
\[\dfrac{{{n}^{2}}-n+2{{n}^{2}}}{2}\]
\[\dfrac{3{{n}^{2}}-n}{2}\]

Hence, we can say that the total number of ways of selecting 2 numbers such that their sum is divisible by 3 is \[\dfrac{3{{n}^{2}}-n}{2}\] for set {1, 2, 3, …., 3n}.
Therefore, option (b) is the correct answer.

Note: While solving this question, the possible mistake one can make is by considering 3 cases, which are both numbers of 3A type or both numbers of (3A – 1) type of both numbers of (3A – 2) type which is the totally wrong way to solve. The only possible cases are both numbers of 3A type and one of (3A – 1) type and another (3A – 2) type.