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Question
How many code symbols can be formed using 5 out of 6 letters A, B, C, D, E, F so that the letters
- cannot be repeated
- can be repeated
- cannot be repeated but must begin with E
- cannot be repeated but end with CAB.
Sum
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Solution
Given letters are A, B, C, D, E and F
a) cannot be repeated
- First box can be filled up in 6 ways.
- Second box can be filled up in 5 ways.
- Third box can be filled up in 4 ways.
- Fourth box can be filled up in 3 ways.
- Fifth box can be filled up in 2 ways.
∴ By fundamental principle of multiplication, total number of code symbols = 6 × 5 × 4 × 3 × 2 = 720.
b) can be repeated
Since the letters can be repeated, all the 5 boxes can be filled up in 6 × 6 × 6 × 6 × 6 ways = 7776.
c) cannot be repeated but must begin with E
| E |
- Since the letters cannot be repeated,
- Second box can be filled up in 5 ways.
- Third box can be filled up in 4 ways.
- Fourth box can be filled up in 3 ways.
- Fifth box can be filled up in 2 ways.
∴ Total number of code symbols = 5 × 4 × 3 × 2 = 120.
d) cannot be repeated but end with CAB.
| C | A | B |
- Since the letters cannot be repeated,
- I box can be filled up in 3 ways.
- II box can be filled up in 2 ways.
∴ Total number of code symbols = 3 × 2 = 6.
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Chapter 2: Algebra - Miscellaneous Problems [Page 47]
