CAT 2003 Slot 2 — QA Question 40
Answer the following question based on the information given below.
A string of three English letters is formed as per the following rules:
- The first letter is any vowel.
- The second letter is m, n or p.
- If the second letter is m, then the third letter is any vowel which is different from the first letter.
- If the second letter is n, then the third letter is e or u.
- If the second letter is p, then the third letter is the same as the first letter.
An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, ... , 9 such that the first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create confusion, when read upside down – for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?
Answer & solution
- A
80
- B
78
71
- D
69
From the digits 0, 1, 2, … , 9, there are 4 digits that can create confusion are 1, 6, 8 and 9
Numbers with 0 in unit’s place cannot be counted because reversing them will give an invalid code.
Codes for which confusion can arise:
Possible number of digits in ten’s place = 4
Possible number of digits in unit’s place, such that the 2 digits are distinct = 3
Total number of ways = 4 × 3 = 12
However, the numbers 69 and 96 do not create confusion when written upside down.
∴ Total number of ways for codes for which confusion can arise = 12 − 2 = 10
Total codes possible with digits from 0 to 9:
Possible number of digits in ten’s place = 9 (0 cannot be used)
Possible number of digits in one’s place, such that the 2 digits are distinct = 9
Total number of all possible codes = 9 × 9 = 81
∴ Number of codes with no confusion = 81 – 10 = 71
Hence, option (c).