Ans.

(i)

Sol. 3n natural numbers can be divided into three groups as :

G 1 : 3, 6, 9, ....... 3n

G 2 : 1, 4, 7, ....... 3n – 2

G 3 : 2, 5, 8, ........ 3n – 1

Each group has n elements.

(a + b + c) is divisible by 3 if

Case (i)

All three integers are from same group

No. of ways of selection = 3. n C 3

Case (ii)

One integer from each group

No. of ways of selection = n C 1 . n C 1 . n C 1 = n 3 ∴ Total ways = 3

n C 3 + n 3 = (3n

2 – 3n + 2)

(ii)

Sol. 3n natural numbers can be divided into three groups as :

G 1 : 3, 6, 9, ....... 3n

G 2 : 1, 4, 7, ....... 3n – 2

G 3 : 2, 5, 8, ........ 3n – 1

Each group has n elements.

(a 2 – b 2 ) is divisible by 3 if either (a – b) or (a + b) is divisible by 3 which is possible if

Case (i)

Both are chosen from same group = 3 . n C 2

Case (ii)

One is chosen from Group G 2 and other from G 3 = n C 1 .

n C 1 ∴ Total ways = 3 n C 2 + n 2 = + n(iii)

Sol. 3n natural numbers can be divided into three groups as :

G 1 : 3, 6, 9, ....... 3n

G 2 : 1, 4, 7, ....... 3n – 2

G 3 : 2, 5, 8, ........ 3n – 1

Each group has n elements.

(a

3 + b 3 ) is divisible by 3

We know (a 3 + b 3 ) = (a + b) 3 – 3ab(a + b)

So, if (a + b) is divisible by 3, then a 3 + b 3 will be divisible by 3

Following cases possible

(i) when a, b are from group G 1

No. of ways of selection = n C 2

(ii) one is from group G 2 and other from G 3

No. of ways of selection = n C 1 . n C 1 = n 2 ∴ Total ways =

n C 2 + n 2 =