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Difficulty: HardPermutations and Combinations

A 5-digit passcode is to be created using five distinct digits chosen from the set {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}. The passcode must include both the digit 33 and the digit 55, and the digit 33 must appear somewhere to the left of the digit 55 in the passcode. How many such 5-digit passcodes can be formed?

Answer: 600 passcodes

Answer

600
To form a valid passcode, first select 3 digits from the 5 available digits {1,2,4,6,7}\{1, 2, 4, 6, 7\}, which can be done in (53)=10\binom{5}{3} = 10 ways. Each chosen set of 5 distinct digits (including 3 and 5) can be arranged in 5!=1205! = 120 ways. Because the digits 3 and 5 are distinct, the digit 3 appears before the digit 5 in exactly half of these arrangements (120/2=60120 / 2 = 60). Multiplying the 10 combinations of digits by the 60 valid arrangements gives a total of 600600 valid passcodes.

Step-by-Step Solution

1
Determine the number of ways to choose the remaining digits.
(53)=10\binom{5}{3} = 10 ways to choose 3 additional digits from {1,2,4,6,7}\{1, 2, 4, 6, 7\}.
Since the passcode must contain both 3 and 5, 3 additional distinct digits must be selected from the 5 available remaining digits.
2
Calculate the total permutations of the 5 chosen digits.
5!=1205! = 120 total permutations.
Any set of 5 distinct digits can be arranged into a 5-digit sequence in 5!5! ways.
3
Apply the positional restriction using symmetry.
1202=60\frac{120}{2} = 60 valid arrangements per set of digits.
In exactly half of all permutations containing both 3 and 5, the digit 3 appears before the digit 5.
4
Compute the total number of valid passcodes.
10×60=60010 \times 60 = 600 passcodes.
Multiply the number of digit selections by the number of valid orderings per selection.

Key Concept

Combining selection (combinations) with symmetry-restricted arrangements (permutations)
Estimated Time:2m 0s
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Topics

Permutations and Combinations

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