Answer: (a)

If we fix 1 at the unit’s place, the other digits can be arranged in 4! = 24 ways.

So, there are 24 numbers which have 1 at the unit’s place and sum of these numbers is 24.

Similarly, there will be 24 numbers each with 3, 5, 7 and 9 at the unit’s place.

So, the sum of all the numbers at the unit’s place will be:

24 + 24 × 3 + 24 × 5 + 24 × 7 + 24 × 9 = 24 + 72 + 120 + 168 + 216 = 600.

Similarly, the sum of the digits at the ten’s place, hundred’s place,

thousand’s place and ten thousand’s place will be 600.

So, the sum of all numbers = 600 (1 + 10 +100 +1,000 +10,000) = 6666600.

Alternative method:

The sum of all the numbers formed by the digits a₁, a₂, a₃,……….a_n, without repetition of the digits is given by:

(n-1)!(a₁ + a₂ + a₃ + ………a_n) (10ⁿ -1)/9

Hence, the sum of the given numbers = 4!(1 + 3 + 5 + 7 + 9) × 11111 = 6666600.

Answer: (b)

Out of the 5 girls,3 girls are to be invited. It can be done in ways = 10 ways.

Now, nothing is given about the number of boys to be invited.

He can invite one boy, two boys, three boys, all the four boys or no boy.

Hence, the total number of ways of selection of boys = 2⁴ = 16.

So, the total number of ways of invitation = 10 × 16 = 160.

Answer: (b)

According to the question, the digit in the unit’s place should be greater than that in the ten’s place.

Hence, if 5 is placed at the unit place, then remaining four places can be filled with any of the

four digits 1, 2, 3 and 4, hence total numbers =4!

However, if digit 4 is placed at the unit place then 5 cannot occupy the ten’s place.

Hence, the digits at the ten’s place can be 1, 2 or 3. This can happen in 3 ways.

The remaining 3 digits can be filled in the remaining three places in 3! ways.

Hence, in all we have (3 × 3!) numbers ending in 4.

Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2.

This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways.

Hence, we will have (2 × 3!) numbers ending in 3. Similarly, we have 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers=  4! + 3 × 3! + 2 × 3! + 3! = 24 + 18 + 12 + 6 = 60.

Answer: (d)

Since the number is greater than 0 and less than a million, so all the numbers of single digit,

two digits, three digits, four digits, five digits and six digits formed by the digits 0, 7 and 8 will be considered.

Now, the number of ways for selecting single digit = 2

Number of ways for selecting two digit = 2 × 3 = 6

Number of ways for selecting three digits = 2 × 3 × 3 = 18

Number of ways for selecting four digits = 2 × 3 × 3 × 3 = 54

Number of ways for selecting five digits = 2 × 3 × 3 × 3 × 3 = 162

Number of ways for selecting six digits= 2 × 3 × 3 × 3 × 3 × 3= 486

Hence, the total number of ways= (2 + 6 +18 + 54 +162 + 486) = 728.

Answer: (c)

The available digits are 0,1,2, …9.

The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digits repetition not allowed).

Thus, the code can be made in 9 × 9 = 81 ways.

Now there are only 4 digits 1, 6, 8, 9 which can create confusion.

Hence, the total number of codes which create confusion are = 4 × 3 = 12.

Out of these 12 codes 69 and 96 will not create confusion.

Hence, in total 12 – 2 = 10 codes will create confusion.

Hence, the total codes without confusion are 81 – 10 = 71.