Eight students A, B, C, D, E, F, G and H sit around a circular table, equidistant from each other, facing the centre of the table, not necessarily in the same order. B and D sit neither adjacent to C nor opposite to C. A sits in between E and D and F sits in between B and H. Which one of the following is definitely correct?
UPSC - 2022
B sits in between A and G
C sits opposite to G
E sits opposite to F
None of the above
Three persons A, B and C are standing in a queue not necessarily in the same order. There are 4 persons between A and B, and 7 persons between B and C. If there are 11 persons ahead of C and 13 behind A, what could be the minimum number of persons in the queue?
UPSC - 2022
To find the minimum number of persons in the queue, we need to arrange A, B, and C such that there is maximum overlap between the given conditions. This usually means placing them in an order that minimizes the total count, often by having one person between two others, or by considering the relative positions from the front and back of the queue.
There are two primary arrangements to consider for A, B, and C to minimize the total number of people:
This arrangement implies that B is between A and C. This is not optimal for minimum as it would mean C is further back than A. Let's check the other case.
This arrangement allows for overlap in the counts. Let's place A, B, and C in this order to minimize the total. For this to happen, C must be between A and B.
Using the position of A from the front (\(P_A = 15\)) and the number of persons behind A (13):
\[ \text{Total persons} = P_A + \text{persons behind A} = 15 + 13 = 28 \]If there are 28 persons in total, and C is the 12th person from the front, then the number of persons ahead of C is \(12 - 1 = 11\). This matches the given information.
If C is before A, and A is before B. This would mean A is after C, and B is after A. This would lead to a larger number of people as A and B are further apart. This is not the minimum case.
If B is before A, and A is before C. This could also be a candidate for minimum. Let's check.
This arrangement (B-A-C) gives a total of 22 persons.
The minimum number of persons is 22.
A) 22 is the minimum number of persons in the queue when the arrangement is B-A-C.
The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
UPSC - 2022
We need to arrange five distinct letters (A, B, C, D, E) such that there are exactly two letters between A and E. This problem can be solved by first treating A and E, along with the two letters between them, as a single block. Then, we arrange the letters within this block and finally arrange the block itself with the remaining letter.
Identify the letters: We have five letters: A, B, C, D, E.
Condition: Exactly two letters must be between A and E. This means we have a structure like A _ _ E or E _ _ A.
Choose the two letters between A and E: The remaining letters are B, C, D. We need to choose 2 letters out of these 3 to place between A and E. The number of ways to choose 2 letters from 3 is given by the combination formula:
\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \times 1} = 3 \]Arrange the two chosen letters: Once the two letters are chosen, they can be arranged in \(2!\) ways. For example, if B and C are chosen, they can be arranged as BC or CB.
\[ 2! = 2 \times 1 = 2 \]Form the block: The block will be of the form (A _ _ E) or (E _ _ A). The number of ways to arrange A and E themselves is \(2!\) (A_ _E or E_ _A).
\[ 2! = 2 \times 1 = 2 \]Total arrangements for the block: Multiply the possibilities from steps 3, 4, and 5 to find the total number of ways to form the block with A and E separated by two letters:
\[ \text{Number of ways to form the block} = \binom{3}{2} \times 2! \times 2! = 3 \times 2 \times 2 = 12 \]Arrange the block and the remaining letter: After forming the block (e.g., A B C E), we have one letter remaining (e.g., D). Now we have two entities to arrange: the block and the remaining letter. Treat the block as a single unit. So, we are arranging 2 items (the block and the remaining letter).
\[ \text{Number of ways to arrange these 2 items} = 2! = 2 \times 1 = 2 \]Total possible arrangements: Multiply the number of ways to form the block by the number of ways to arrange the block and the remaining letter.
\[ \text{Total arrangements} = 12 \times 2 = 24 \]C) 24
A) 12: This would be the answer if we only considered forming the block (A _ _ E) or (E _ _ A) but forgot to arrange this block with the fifth letter. Or if we only considered fixing A and E and arranging the middle letters, but not arranging the block with the remaining letter.
B) 18: This value does not arise from the correct calculation steps. It might be a result of miscalculation or an incorrect understanding of permutations/combinations.
D) 36: This would be the answer if, for example, we incorrectly calculated the arrangements within the block or the arrangement of the block with the remaining letter. For instance, if one thought there were 3 ways to choose the letters, 2 ways to arrange them, and 3 ways to arrange A and E, and then 2 ways to arrange the block and the remaining letter, it would be \(3 \times 2 \times 3 \times 2 = 36\), which is incorrect because A and E can only be arranged in 2 ways.
If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?
UPSC - 2017
This problem involves permutations with restrictions. We need to arrange 2 boys (B) and 2 girls (G) in a row such that the two girls are not next to each other. We can solve this by first finding the total number of arrangements without any restrictions and then subtracting the arrangements where the girls are together.
We have 4 distinct positions for 2 boys and 2 girls. If all 4 people were distinct, the total number of arrangements would be \(4!\). However, the two boys are identical in terms of their 'boy' identity for arrangement purposes, and similarly for the girls. But in permutation problems like this, individuals are usually considered distinct unless specified. Let's assume the boys are \(B_1, B_2\) and girls are \(G_1, G_2\). The total number of ways to arrange 4 distinct people is:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]Treat the two girls (\(G_1G_2\)) as a single unit. Now we are arranging 3 units: \(B_1, B_2\), and \((G_1G_2)\). The number of ways to arrange these 3 units is:
\[ 3! = 3 \times 2 \times 1 = 6 \]Within the unit of girls \((G_1G_2)\), the girls can also be arranged in \(2!\) ways (i.e., \(G_1G_2\) or \(G_2G_1\)).
\[ 2! = 2 \times 1 = 2 \]So, the total number of arrangements where the girls are together is the product of the arrangements of the units and the arrangements within the girls' unit:
\[ \text{Arrangements (girls together)} = 3! \times 2! = 6 \times 2 = 12 \]Subtract the arrangements where the girls are together from the total number of arrangements:
\[ \text{Arrangements (girls not together)} = \text{Total arrangements} - \text{Arrangements (girls together)} \] \[ = 24 - 12 = 12 \]C) 12
Five people A, B, C, D and E are seated about a round table. Every chair is spaced equidistant from adjacent chairs.
(i) C is seated next to A
(ii) A is seated two seats from D
(iii) B is not seated next to A
On the basis of above information, which of the following must be true?
1. D is seated next to B
2. E is seated next to A
3. D and C are separated by two seats
Select the correct answer using the code given below:
UPSC - 2016
Consider the following: A, B, C, D, E, F, G and H are standing in a row facing North.
B is not neighbour of G.
F is to the immediate right of G and neighbour of E.
G is not at the extreme end.
A is sixth to the left of E.
H is sixth to the right of C.
Which one of the following is correct in respect of the above?
UPSC - 2017
C is to the immediate left of A.
D is immediate neighbour of B and F.
G is to the immediate right of D.
A and E are at the extreme ends.
This problem involves arranging eight people (A, B, C, D, E, F, G, H) in a row facing North based on given relational clues. We will deduce their positions step-by-step to form the final arrangement and then check which option is correct.
This means the sequence is G F E. (G is to the left of F, F is to the left of E).
This means G cannot be at position 1 or 8.
If E is at position \(x\), then A is at position \(x-6\). This implies there are 5 people between A and E. The minimum position for E would be 7 (so A is at 1). The maximum position for A would be 2 (so E is at 8).
Combining with GFE: If we place GFE, the minimum length is 3. If E is at 7, then F is at 6, G is at 5. If A is 6th to the left of E (position 7), A would be at position \(7-6=1\).
So, a possible partial arrangement is: A _ _ G F E _ _
If C is at position \(y\), then H is at position \(y+6\). This implies there are 5 people between C and H. The minimum position for C would be 1 (so H is at 7). The maximum position for C would be 2 (so H is at 8).
Let's try to fit A and E. If A is at 1, E is at 7. This gives:
1 2 3 4 5 6 7 8
A _ _ _ _ _ E _
Now, place GFE (from Clue 1) such that E is at 7. This means F is at 6 and G is at 5.
1 2 3 4 5 6 7 8
A _ _ _ G F E _
This satisfies "G is not at the extreme end" (G is at 5).
Now place C and H (Clue 4). H is 6th to the right of C. The only remaining slots are 2, 3, 4, 8. If C is at 2, H would be at \(2+6=8\).
This gives the arrangement:
1 2 3 4 5 6 7 8
A C _ _ G F E H
The remaining person is D. The only empty slot is 4. So D must be at 4.
The arrangement becomes:
1 2 3 4 5 6 7 8
A C B D G F E H (Wait, B is not placed yet. Let's re-evaluate the C and H placement.)
If C is at 2, H is at 8. The remaining slots are 3 and 4. The remaining people are B and D.
So, the arrangement is A C (B/D) (D/B) G F E H.
Now, apply "B is not neighbour of G". G is at 5. Its neighbours are D (at 4) and F (at 6). This means B cannot be at 4 or 6. Since F is already at 6, B cannot be at 4.
Therefore, D must be at 4. This leaves B for position 3.
The final arrangement is:
1 2 3 4 5 6 7 8
A C B D G F E H
Let's check all conditions:
From our arrangement (A C B D G F E H), C is to the immediate right of A. So, this is incorrect.
From our arrangement, D is at 4. Its neighbours are B (at 3) and G (at 5). It is not a neighbour of F (F is at 6). So, this is incorrect.
From our arrangement, D is at 4 and G is at 5. G is indeed to the immediate right of D. So, this is correct.
From our arrangement, A is at the left extreme end (position 1), but E is at position 7, not the right extreme end (position 8, which is H). So, this is incorrect.
Correct Option: C) G is to the immediate right of D.
