## Previous Year Paper

### CAT-2005-Previous Years Paper

Question
15 out of 30

Let S be the set of five-digit numbers formed by digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd position are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

 A 228 B 216 C 294 D 192

Ans. B

Odd positions can be counted in 2 ways.

(i) Counting form the LMD – end:

We have 1, 2, 3, 4, and 5 to be filled in these blocks. Odd numbers (1, 3, 5) to be be filled at odd positions. Other places are to be filled by even numbers (2 or 4). Let us count, how many such numbers are there with 2 at the unit’s digit.

Odd numbers can be filled 3P2 = 6 ways

The remaining two places are to be filled by 2 numbers (one odd number left out of 1, 3, 5 and one even i.e., 4) in = 2 ways

S, there are 6 × 2 = 12 number with 2 at the right most place. Similarly there are 12 such numbers with 4 at the rightmost digits.

The sum of the rightmost digits in all such number = 12 (2 + 4) = 72

(ii) Now counting from the RMD-end.

Let us place 1 at the unit’s place and check, how many numbers are possible with (1, 3) at the odd positions:

Number of such cases = 2 × 2 = 4 ways

Here again number of ways = 2 × 2 = 4 ways

So, there are 4 + 4 = 8 numbers in which (1, 3) are at odd positions. Similarly there are 8 numbers in which (1, 5) are at odd positions. So, in all there are 16 numbers, where 1 is at the unit’s place. Similarly there are 16 numbers with 3 at unit’s place and 16 more with 5 at the unit’s place.

Summing up all the odd unit’s digits = 16 (1 + 3 + 5) = 144

From (i) and (ii) we can now sum all (even or odd) numbers at unit place = 72 + 144 = 216