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If the Board of Selectmen contains 4 positions, and if in the current 12 May 2023, 03:10
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73% (00:54) correct 27% (01:22) wrong based on 74 sessions

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If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, then in how many different ways can these candidates be elected to the Board?

(A) 6
(B) 8
(C) 12
(D) 16
(E) 24
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D
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If the Board of Selectmen contains 4 positions, and if in the current 12 May 2023, 22:17
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Bunuel
If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, then in how many different ways can these candidates be elected to the Board?

(A) 6
(B) 8
(C) 12
(D) 16
(E) 24
Show more

For each position on the board, we can have one of the two candidates elected. Therefore each position can be 'filled' in two ways. As there are four positions the number of ways to fill the four positions =

\(2 * 2 * 2 * 2 = 2^4 = 16\)

Option D
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If the Board of Selectmen contains 4 positions, and if in the current 29 Dec 2023, 08:13
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gmatophobia
Bunuel
If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, then in how many different ways can these candidates be elected to the Board?

(A) 6
(B) 8
(C) 12
(D) 16
(E) 24

For each position on the board, we can have one of the two candidates elected. Therefore each position can be 'filled' in two ways. As there are four positions the number of ways to fill the four positions =

\(2 * 2 * 2 * 2 = 2^4 = 16\)

Option D
Show more

I do not conceptually understand this explanation. To me, the answer is not 2^4 but 4^2, the conceptual difference being that either of the two candidates can fill either of the four slots, thus, 4*4. The quoted explanation, however, says either of the four position can be filled by either of the two candidates but I do not think that is the case. Only two slots can be claimed by the two candidates, not all four.

Bunuel can you please help me understand?

EDIT: I thought the answer would be 4C2. Therefore, 4!/(2!2!) = 3*2 = 6. I got the wrong answer due to a conceptual gap.
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If the Board of Selectmen contains 4 positions, and if in the current 29 Dec 2023, 08:25
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Khwarizmi
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Bunuel
If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, then in how many different ways can these candidates be elected to the Board?

(A) 6
(B) 8
(C) 12
(D) 16
(E) 24

For each position on the board, we can have one of the two candidates elected. Therefore each position can be 'filled' in two ways. As there are four positions the number of ways to fill the four positions =

\(2 * 2 * 2 * 2 = 2^4 = 16\)

Option D

I do not conceptually understand this explanation. To me, the answer is not 2^4 but 4^2, the conceptual difference being that either of the two candidates can fill either of the four slots, thus, 4*4. The quoted explanation, however, says either of the four position can be filled by either of the two candidates but I do not think that is the case. Only two slots can be claimed by the two candidates, not all four.

Bunuel can you please help me understand?

EDIT: I thought the answer would be 4C2. Therefore, 4!/(2!2!) = 3*2 = 6. I got the wrong answer due to a conceptual gap.
Show more

There are 4 positions to be filled, and for each position, there are two candidates competing:

    A and B are running for the first slot.
    C and D are running for the second slot.
    E and F are running for the third slot.
    G and H are running for the fourth slot.

To determine the total number of ways these candidates can be elected to the Board, we simply multiply the options for each slot together:

Total ways = 2 (options for the first slot) * 2 (options for the second slot) * 2 (options for the third slot) * 2 (options for the fourth slot) = 2^4 = 16

Does this make sense?
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If the Board of Selectmen contains 4 positions, and if in the current 29 Dec 2023, 10:40
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Bunuel
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gmatophobia


For each position on the board, we can have one of the two candidates elected. Therefore each position can be 'filled' in two ways. As there are four positions the number of ways to fill the four positions =

\(2 * 2 * 2 * 2 = 2^4 = 16\)

Option D

I do not conceptually understand this explanation. To me, the answer is not 2^4 but 4^2, the conceptual difference being that either of the two candidates can fill either of the four slots, thus, 4*4. The quoted explanation, however, says either of the four position can be filled by either of the two candidates but I do not think that is the case. Only two slots can be claimed by the two candidates, not all four.

Bunuel can you please help me understand?

EDIT: I thought the answer would be 4C2. Therefore, 4!/(2!2!) = 3*2 = 6. I got the wrong answer due to a conceptual gap.

There are 4 positions to be filled, and for each position, there are two candidates competing:

    A and B are running for the first slot.
    C and D are running for the second slot.
    E and F are running for the third slot.
    G and H are running for the fourth slot.

To determine the total number of ways these candidates can be elected to the Board, we simply multiply the options for each slot together:

Total ways = 2 (options for the first slot) * 2 (options for the second slot) * 2 (options for the third slot) * 2 (options for the fourth slot) = 2^4 = 16

Does this make sense?
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It does. Thanks a lot! I misunderstood the question, thinking two people, A and B, were competing over four positions.
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If the Board of Selectmen contains 4 positions, and if in the current 30 Dec 2023, 09:12
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If the Board of Selectmen contains 4 positions, and if in the current election two candidates are running for each position, then in how many different ways can these candidates be elected to the Board?
Sol=> for each of the 4 positions the position can be filled in two positions each so ,
2*2*2*2 = 2^4 ways
= 16
hence option D
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