NIMCET 2027 Permutations And Combinations Previous Year Questions (PYQs) – Page 2 of 5

Total bottles and boxes: 9 each, labeled 1 to 9.

We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.

Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).

Number of ways = $\binom{9}{5}$

Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).

Let $D_4$ be the number of derangements of 4 items.

$D_4 = 9$

Step 3: Total ways = $\binom{9}{5} \times D_4 = 126 \times 9 = 1134$

✅ Final Answer: $\boxed{1134}$

Question: Find the value of:

$$ \sum_{r=1}^{n} \frac{nP_r}{r!} $$

Solution:

We know: \( nP_r = \frac{n!}{(n - r)!} \Rightarrow \frac{nP_r}{r!} = \frac{n!}{(n - r)! \cdot r!} = \binom{n}{r} \)

Therefore,

$$ \sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} \binom{n}{r} = 2^n - 1 $$

Final Answer: $$ \boxed{2^n - 1} $$

Total Number of Intersection Points

Given:

• 10 distinct lines: \( L_1, L_2, \ldots, L_{10} \)
• \( L_2, L_4, L_6, L_8, L_{10} \): parallel (no intersections among them)
• \( L_1, L_3, L_5, L_7, L_9 \): concurrent at point \( C \) (intersect at one point)

? Calculation:

\[ \text{Total line pairs: } \binom{10}{2} = 45 \]

\[ \text{Subtract parallel pairs: } \binom{5}{2} = 10 \Rightarrow 45 - 10 = 35 \]

\[ \text{Concurrent at one point: reduce } 10 \text{ pairs to 1 point} \Rightarrow 35 - 9 = \boxed{26} \]

✅ Final Answer: \(\boxed{26}\) unique points of intersection