A 6-digit security code is to be formed using distinct digits chosen from the set \{1, 2, … — GRE General Test

A 6-digit security code is to be formed using distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ . The code must satisfy the following conditions:
1. The code must be an even number (its final digit must be $2$ , $4$ , $6$ , or $8$ ).
2. Both digits $1$ and $2$ must be included in the 6-digit code.
3. Digits $1$ and $2$ cannot occupy adjacent positions in the code.

Which of the following statements regarding the number of possible 6-digit security codes must be true? Select all such statements.

The total number of valid 6-digit security codes that satisfy all conditions is $3,600$ .Cevap
The number of valid security codes in which the final digit is $2$ is $1,440$ .Cevap
The number of valid security codes in which the final digit is an even digit other than $2$ is $2,160$ .Cevap
D
If the non-adjacency constraint between $1$ and $2$ were removed, the total number of valid security codes would be $4,320$ .
E
There are $20$ valid ways to place digits $1$ and $2$ non-adjacently in the first 5 positions of the code.

Cevap

The statements confirming that the total number of codes is 3,600, that 1,440 codes end in 2, and that 2,160 codes end in an even digit other than 2 are all correct.

The solution requires partitioning into two mutually exclusive scenarios depending on whether digit 2 occupies the final position. When digit 2 is at the end, digit 1 can occupy any of the first 4 positions (excluding position 5 to avoid adjacency), yielding

4 \times P(6,4) = 1,440

codes. When the final position is occupied by 4, 6, or 8 (3 choices), digits 1 and 2 have

P(5,2) - 8 = 12

valid non-adjacent placements across the first 5 positions, and the remaining 3 positions can be filled in

P(5,3) = 60

ways, yielding

3 \times 12 \times 60 = 2,160

codes. The sum of these two cases gives

3,600

total valid codes. Thus, the three statements asserting total codes of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are all correct.

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Analyze the conditions and split into two disjoint cases based on the last digit.

Case 1: The last digit (6th position) is

2

. Case 2: The last digit (6th position) is

4

6

, or

8

Digit

2

plays a dual role: it satisfies the even-ending condition and is one of the restricted digits.

Calculate Case 1 (last digit is 2).

Position 6 is fixed as

2

(

1

choice). Digit

1

must be in the first 5 positions but cannot be adjacent to position 6 (so position 5 is excluded). Thus digit

1

has

4

choices (positions 1, 2, 3, 4). The remaining

4

positions are filled from the remaining

6

available digits

\{3, 4, 5, 6, 7, 8\}

P(6, 4) = 6 \times 5 \times 4 \times 3 = 360

ways. Total for Case 1:

1 \times 4 \times 360 = 1,440

This determines the valid arrangements when 2 is forced to the end of the code.

Calculate Case 2 (last digit is 4, 6, or 8).

There are

3

choices for the 6th position. Digits

1

and

2

must be placed among the first 5 positions non-adjacently. Total ordered placements of

1

and

2

in 5 positions is

P(5, 2) = 20

. The number of adjacent position pairs is

4

(positions (1,2), (2,3), (3,4), (4,5)), with

2! = 2

orderings per pair, giving

4 \times 2 = 8

adjacent placements. Non-adjacent placements =

20 - 8 = 12

. The remaining 3 open positions are filled from the remaining 5 available digits in

P(5, 3) = 60

ways. Total for Case 2:

3 \times 12 \times 60 = 2,160

This accounts for codes ending in 4, 6, or 8 while respecting the non-adjacency of 1 and 2.

Combine the cases and verify statements.

Total valid codes =

1,440 + 2,160 = 3,600

. The statements asserting totals of 3,600, 1,440 ending in 2, and 2,160 ending in 4, 6, or 8 are true.

Summing mutually exclusive cases gives the overall number of valid outcomes.

Anahtar Kavram

Fundamental Counting Principle with Permutations under Restricted Adjacency

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