Q.If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

We are told that $AX = 0$ is a homogeneous system of three equations in three unknowns, so $A$ is a $3 \times 3$ matrix. The vector

represents an infinite number of solutions. That means there is not just one isolated solution; there is a whole family of solutions parameterized by $l$ and $m$ (both free to vary, except they cannot be zero simultaneously? Actually the statement says $l \neq 0$, $m \neq 0$, but that might just be to avoid trivial zero vector; the key is that $l$ and $m$ can vary independently, giving infinitely many distinct solutions).

Key concept: For a homogeneous system $AX=0$, the set of all solutions is the nullspace of $A$. The dimension of the nullspace is called the nullity. The rank-nullity theorem says:

So if we can determine the nullity, we immediately get the rank.

Now, the given solution vectors all have the third component equal to 0. That means every solution in this family lies in the $xy$-plane (first two coordinates free, third fixed at 0). So the nullspace contains at least the set

This set is a 2‑dimensional subspace (since $l$ and $m$ are independent parameters). Therefore the nullity is at least 2.

But could the nullity be 3? If nullity = 3, then every vector in $\mathbb{R}^3$ would be a solution, including vectors with a non‑zero third component. However, the problem says that the given form represents an infinite number of solutions — it does not say that all solutions are of that form. But if nullity were 3, then certainly vectors like $\begin{bmatrix}0\\0\\1\end{bmatrix}$ would also be solutions, which is not contradicted by the statement. So nullity could be 3? Let’s check: If nullity = 3, then rank = 0, meaning $A$ is the zero matrix. Then every vector is a solution, including the given family. That is possible in principle. But is it the intended answer? Usually in such multiple‑choice problems, they imply that the given family is the entire solution set or at least that the solution set is exactly that 2‑dimensional plane. The phrasing “represents an infinite number of solutions” is ambiguous, but the typical interpretation in exam problems is: the set of all solutions is exactly that family (or at least that the family shows the nullspace is at least 2‑dimensional, and we need to decide if it could be larger). Let’s examine the options: rank = 3 would mean only the trivial solution, which contradicts having any non‑zero solution. So (A) is out. Rank = 2 gives nullity = 1, meaning only one free parameter — but here we have two independent parameters $l$ and $m$, so nullity cannot be 1. So (B) is out. Rank = 1 gives nullity = 2, which matches exactly two free parameters. Rank = 0 gives nullity = 3, which would also allow two free parameters (indeed three), but then the rank would be 0, which is not listed as an option — (D) says “does not exist”, but rank always exists for a matrix; it’s a non‑negative integer. So (D) is not correct. Therefore the only plausible answer is rank = 1.

But let’s be thorough: Could the nullity be 3? That would mean $A$ is the zero matrix. Then the system is $0=0$, and every vector is a solution. The given family is a subset of all solutions, so it’s still true that it “represents an infinite number of solutions”. However, the problem likely intends that the solution set is exactly that 2‑dimensional plane, not the whole space. Moreover, if $A$ were zero, the rank would be 0, which is not among the choices (A, B, C). Option (D) “does not exist” is a distractor; rank always exists. So the intended answer is rank = 1.

Thus we conclude:

The system $AX=0$ is homogeneous in three unknowns, so $A$ is a $3\times3$ matrix. Its solution set is

Because $l$ and $m$ range independently over $\mathbb{R}$, the solution (null) space is spanned by the two independent vectors $[1,0,0]^{T}$ and $[0,1,0]^{T}$. Hence

By the rank–nullity theorem for the $3$ columns of $A$:

Concept & Intuition

The problem imposes a strict spacing rule: between any two girls, there must be exactly two boys. This is not a flexible “at least two” condition — it’s exact. That means the positions of the girls relative to the boys are forced into a single repeating pattern. Once we fix that pattern, we just count the permutations of the girls among the girl positions and the boys among the boy positions. The trap is to think we can place the girls arbitrarily and then insert boys; but the exact spacing locks the structure completely.

Step-by-step reasoning

1. Understand the constraint We have 4 girls. Between any two consecutive girls, there must be exactly 2 boys. That means the arrangement must look like:

This uses $4$ girls and $3 \times 2 = 6$ boys — exactly all the boys we have. So the pattern is forced: there is no room for extra boys at the ends or anywhere else.

2. Check if any other pattern is possible

   Could we put boys before the first girl or after the last girl? If we did, then between the first and second girl we would still need exactly two boys, but the total number of boys would exceed 6. Since we have exactly 6 boys, the only way to satisfy the “exactly two between every pair” condition is to use all boys in the interior gaps, leaving none for the ends. So the pattern above is unique.

3. Count arrangements of the girls

   The 4 girls occupy the 4 positions marked $G$. They can be arranged among themselves in $4!$ ways.

4. Count arrangements of the boys

   The 6 boys occupy the 6 positions marked $B$. They can be arranged among themselves in $6!$ ways.

5. Multiply

   Since the choices for girls and boys are independent, the total number of arrangements is:

6. Match with options

   Option (A) is $6!5!$, which is not the same as $4!6!$. Wait — careful: $4! = 24$ and $5! = 120$, so $6!5! = 720 \times 120 = 86400$, while $4!6! = 24 \times 720 = 17280$. So (A) is not our answer? Let’s re-check: The problem lists (A) as $6!5!$. But our result is $4!6!$. However, note that $4!6! = 6! \times 24$ and $6!5! = 6! \times 120$. So they differ. Did we miss something?

   Watch out

   A common mistake is to think the pattern can start or end with a boy. If we allowed boys at the ends, we would need more than 6 boys to keep exactly two between every pair. For example, G B B G B B G B B uses 6 boys and 4 girls, but if we add a boy at the start, we have 7 boys — not allowed. So the pattern is indeed fixed as above, giving $4!6!$. But option (A) is $6!5!$. Let’s check the options again:

   (A) $6!5!$

   (B) $(72)6!$

   (C) $(144)5!$

   (D) $4!7!$

   Our result $4!6!$ is not listed. So perhaps the intended interpretation is different.

7. Re-examine the problem statement

   “Between any two girls there must be exactly 2 boys” — this could also mean that if you look at any pair of girls (not necessarily consecutive), the number of boys between them is exactly 2. That would be a much stronger condition, forcing all girls to be separated by exactly two boys, which again gives the same pattern. So that doesn’t change.

   Alternatively, maybe the problem means that between any two consecutive girls there are exactly 2 boys, but the row can start or end with boys? Let’s test: Suppose we have B G B B G B B G B B G. That uses 1 boy at start + 6 between = 7 boys, too many. What about G B B G B B G B B G B? That’s 7 boys again. So any extra boy at an end forces total boys > 6. So the only pattern with exactly 6 boys is the one with no boys at ends.

   But then why is $4!6!$ not an option? Let’s compute $4!6! = 24 \times 720 = 17280$. Option (A) $6!5! = 720 \times 120 = 86400$. Option (B) $72 \times 720 = 51840$. Option (C) $144 \times 120 = 17280$. Option (D) $24 \times 5040 = 120960$. So option (C) equals $144 \times 5! = 144 \times 120 = 17280$, which is exactly $4!6!$. So the correct match is (C).

Solution.

1. With $4$ girls in a row there are $3$ gaps between consecutive girls. Each must hold exactly $2$ boys, using $3\times2=6$ boys — all of them.

2. Hence no boys remain for the two ends, so the layout is completely fixed:

3. Count the internal arrangements:

   • the $4$ girls fill the $4$ girl‑slots in $4!$ ways,
   • the $6$ boys fill the $6$ boy‑slots in $6!$ ways.

4. Total:

Concept and intuition:

When we require all boys to be together and all girls to be together, we are essentially treating the entire group of boys as a single "block" and the entire group of girls as another "block". These two blocks can be arranged in $2! = 2$ ways (boys block then girls block, or girls block then boys block). Inside each block, the $n$ boys can be permuted in $n!$ ways, and the $n$ girls in $n!$ ways. So the total number is $2 \times n! \times n! = 2 (n!)^2$.

Now we need to see which of the given options equals this same number.

1. Option (A): The total number of arrangements of $n$ boys and $n$ girls in a row, with no restrictions, is $(2n)!$. This is far larger than $2 (n!)^2$ for $n>1$, so not equal.

2. Option (B): The number of arrangements where all girls are together. Treat the $n$ girls as one block. Then we have $n$ boys + 1 block = $n+1$ objects to arrange, which can be done in $(n+1)!$ ways. Inside the girls' block, the $n$ girls can be permuted in $n!$ ways. So total = $(n+1)! \times n! = (n+1) \times n! \times n! = (n+1) (n!)^2$. This is not the same as $2 (n!)^2$ unless $n+1 = 2$, i.e., $n=1$. For general $n$, they differ. So (B) is not equal.

   Watch out

   A common mistake is to think "all girls together" is symmetric to "all boys together and all girls together". But here we only require one group to be together, not both, so the count is different.

3. Option (C): Arrangements where no two girls are together. For $n$ boys and $n$ girls, the only way to have no two girls together is to place girls in the gaps between boys. With $n$ boys, there are $n+1$ gaps (including ends). We need to choose $n$ of these gaps for the girls, which can be done in $\binom{n+1}{n} = n+1$ ways. Then arrange boys in $n!$ ways and girls in $n!$ ways. Total = $(n+1) \times n! \times n! = (n+1) (n!)^2$. This is the same as option (B), not $2 (n!)^2$.

4. Option (D): Arrangements where no two girls are together and no two boys are together. With equal numbers $n$ and $n$, this forces an alternating pattern. There are exactly two alternating patterns: BGBGBG... or GBGBGB... For each pattern, we arrange the $n$ boys in $n!$ ways and the $n$ girls in $n!$ ways. So total = $2 \times n! \times n! = 2 (n!)^2$. This matches exactly.

Thus the number of arrangements with all boys together and all girls together equals the number of arrangements with no two girls together and no two boys together.

Concept and Intuition

This problem deals with permutations where certain elements have a fixed relative order. When we arrange a set of distinct items, say $n$ items, there are $n!$ ways to do so. However, if we impose a condition that a specific subset of $k$ items must appear in a particular relative order, this significantly reduces the number of possible arrangements.

Consider the three scientists $S_1, S_3,$ and $S_7$. The condition states that $S_1$ must lecture before $S_3$, and $S_3$ must lecture before $S_7$. This means their order must always be $S_1, S_3, S_7$ whenever they appear in the sequence of lectures.

Imagine we have 7 slots for the lectures. If we pick any 3 slots out of these 7 for $S_1, S_3,$ and $S_7$, there is only one way to place them in those 3 slots to satisfy the given condition (i.e., $S_1$ in the earliest slot, $S_3$ in the middle, and $S_7$ in the latest). If there were no such condition, these three scientists could be arranged in $3!$ ways within those 3 chosen slots.

This implies that for every $3!$ arrangements of $S_1, S_3, S_7$ among themselves, only one is valid. Therefore, we effectively "remove" the $3!$ permutations of these three specific scientists from the total number of permutations of all 7 scientists. This is equivalent to treating $S_1, S_3,$ and $S_7$ as if they were identical items for the purpose of arrangement, because their internal ordering is fixed and does not contribute to distinct valid arrangements.

Step-by-Step Solution

1. Identify the total number of items and constrained items:

   We have 7 scientists in total, $S_1, S_2, \ldots, S_7$.

   The lectures of three scientists, $S_1, S_3,$ and $S_7$, have a specific relative order: $S_1$ must be prior to $S_3$, and $S_3$ must be prior to $S_7$. This means their sequence must always be $S_1, S_3, S_7$.

2. Apply the concept of fixed relative order:

   When the relative order of a subset of items is fixed, we can treat those items as indistinguishable (identical) for the purpose of calculating permutations. This is because, once positions are chosen for them, there is only one way to place them to satisfy the fixed order.

   In this case, $S_1, S_3,$ and $S_7$ must appear in the order $S_1 \rightarrow S_3 \rightarrow S_7$. There are 3 such scientists.

3. Calculate the number of arrangements:

   If all 7 scientists were distinct and had no ordering constraints, the number of ways to arrange their lectures would be $7!$.

   However, since the relative order of $S_1, S_3,$ and $S_7$ is fixed, we divide the total permutations by the number of ways these three scientists could arrange themselves, which is $3!$.

   The number of ways to arrange the lectures is given by:

4. Perform the calculation:

The key idea is to treat the dictionary order as alphabetical sorting. We fix letters one by one from the left, and for each position, count how many permutations of the remaining letters would start with a letter that comes earlier in the alphabet. Summing these counts gives the number of words before the target word; adding 1 gives the rank.

Step-by-step reasoning:

1. List the letters of 'HANDLE' in alphabetical order:

   A, D, E, H, L, N.

   The word 'HELAND' has letters H, E, L, A, N, D.

2. Fix the first letter. The first letter of 'HELAND' is H.

   Letters that come before H in the alphabet are A, D, E.

   For each such letter, the remaining 5 letters can be arranged in $5! = 120$ ways.

   So words starting with A, D, or E: $3 \times 120 = 360$ words.

3. Fix the first two letters as 'HE'. Now consider the second letter.

   After H, the next letter in 'HELAND' is E.

   Among the remaining letters (A, D, L, N), which come before E? Only A and D.

   For each such choice (HA or HD), the remaining 4 letters can be arranged in $4! = 24$ ways.

   So words starting with HA or HD: $2 \times 24 = 48$ words.

4. Fix the first three letters as 'HEL'. The third letter is L.

   Remaining letters: A, D, N.

   Letters before L in the alphabet are A and D.

   For each (HEA or HED), the remaining 3 letters arrange in $3! = 6$ ways.

   So words starting with HEA or HED: $2 \times 6 = 12$ words.

5. Fix the first four letters as 'HELA'. The fourth letter is A.

   Remaining letters: D, N.

   Letters before A? None (A is the smallest). So 0 words here.

6. Fix the first five letters as 'HELAN'. The fifth letter is N.

   Remaining letter: D.

   Letters before N? Only D.

   For the prefix HELAD, the last letter is fixed (only one arrangement), so $1! = 1$ word.

   So words starting with HELAD: 1 word.

7. Now sum all words before 'HELAND':

   $360 + 48 + 12 + 0 + 1 = 421$.

8. Rank = number of words before + 1 = $421 + 1 = 422$.

Setup. The multiset is $\{D,D,E,E,I,N\}$; total distinct arrangements $=\dfrac{6!}{2!\,2!}=180$. At each position we count arrangements of the remaining letters that begin with a letter smaller than the one in NIDDEE.

Position 1 (N). Smaller letters D, E, I:

• D first, remaining $\{D,E,E,I,N\}$: $\dfrac{5!}{2!}=60$
• E first, remaining $\{D,D,E,I,N\}$: $\dfrac{5!}{2!}=60$
• I first, remaining $\{D,D,E,E,N\}$: $\dfrac{5!}{2!\,2!}=30$

Subtotal $=150$.

Position 2 (I), prefix N, remaining $\{D,D,E,E,I\}$. Smaller than I: D, E:

• D: remaining $\{D,E,E,I\}$: $\dfrac{4!}{2!}=12$
• E: remaining $\{D,D,E,I\}$: $\dfrac{4!}{2!}=12$

Subtotal $=24$.

Positions 3-5. Prefix NI leaves $\{D,D,E,E\}$; the next letters D, D, E are each the smallest available, so no earlier words arise: $0+0+0$.

Rank.