NIMCET 2027 Permutations And Combinations Previous Year Questions (PYQs) – Page 5 of 5
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Showing 48 questions
If n and r are integers such that 1 ≤ r ≤ n, then the value of n n-1Cr-1is
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Solution
There are 8 students appearing in an examination of which 3 have to appear in Mathematics paper and the remaining 5 in different subjects. Then, the number of ways they can be made to sit in a row, if the candidates in Mathematics cannot sit next to each other is
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Solution
The number of bit strings of length 10 that contain either five consecutive 0’s or five consecutive
1’s is
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Solution
We count bit strings of length 10 that contain at least one run of five identical bits.
Case 1: Exactly one block of five consecutive 0’s.
The block $00000$ can start at positions 1 to 6, so 6 choices.
The remaining 5 positions can be filled freely with 0 or 1, except they should not create another block of five 0’s.
Valid fillings = $2^5 - 1 = 31$.
So number of strings with exactly one block of five 0’s is
$6 \times 31 = 186$
Case 1: Exactly one block of five consecutive 0’s.
The block $00000$ can start at positions 1 to 6, so 6 choices.
The remaining 5 positions can be filled freely with 0 or 1, except they should not create another block of five 0’s.
Valid fillings = $2^5 - 1 = 31$.
So number of strings with exactly one block of five 0’s is
$6 \times 31 = 186$
Case 2: Exactly one block of five consecutive 1’s.
By symmetry, the count is the same.
$186$
By symmetry, the count is the same.
$186$
Case 3: One block of five 0’s and one block of five 1’s.
This is possible only when the blocks do not overlap.
The only such strings are
$0000011111$ and $1111100000$
This is possible only when the blocks do not overlap.
The only such strings are
$0000011111$ and $1111100000$
So total such strings = $2$.
Using inclusion–exclusion principle:
$186 + 186 + 2 = 222$
Final Answer: $222$
Using inclusion–exclusion principle:
$186 + 186 + 2 = 222$
Final Answer: $222$
In an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is
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Solution
Total Papers: 9
Condition for Success: Passes > Fails
So, candidate is unsuccessful when: Passes ≤ 4
Calculate ways:
\[ \text{Ways} = \sum_{x=0}^{4} \binom{9}{x} = \binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4} \]\[= 1 + 9 + 36 + 84 + 126 = \boxed{256}\]
✅ Final Answer: 256 ways
There are $4$ books on fairy tales, $5$ novels and $3$ plays. In how many ways can they be arranged in the order books on fairy tales, novels and then plays so that books of same category are put together
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Solution
Given:
$4$ books on fairy tales, $5$ novels, $3$ plays
The order of arrangement is fixed as:
Fairy Tales $\rightarrow$ Novels $\rightarrow$ Plays
Step 1: Arrange books within each category
Fairy tales books can be arranged among themselves $= 4!= 24$ ways
Novels can be arranged among themselves $= 5! = 120$ ways
Plays can be arranged among themselves $= 3! = 6$ ways
Step 2: Total number of arrangements
$= 4! \times 5! \times 3!$
$= 24 \times 120 \times 6$
$\therefore \boxed{= 17280 \text{ ways}}$
$4$ books on fairy tales, $5$ novels, $3$ plays
The order of arrangement is fixed as:
Fairy Tales $\rightarrow$ Novels $\rightarrow$ Plays
Step 1: Arrange books within each category
Fairy tales books can be arranged among themselves $= 4!= 24$ ways
Novels can be arranged among themselves $= 5! = 120$ ways
Plays can be arranged among themselves $= 3! = 6$ ways
Step 2: Total number of arrangements
$= 4! \times 5! \times 3!$
$= 24 \times 120 \times 6$
$\therefore \boxed{= 17280 \text{ ways}}$
How many even integers between 4000 and 7000 have four different digits?
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Solution
1: Thousands digit
The number is 4-digit and must be between 4000 and 7000.
So the thousands digit can be: 4, 5, or 6.
Step 2: Units digit
The number must be even, so the units digit must be one of {0, 2, 4, 6, 8}.
3: Count cases
- Case 1: Thousands digit = 4
Units digit ≠ 4 → possible units = {0, 2, 6, 8} → 4 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 4 × 8 × 7 = 224. - Case 2: Thousands digit = 5
Units digit can be {0, 2, 4, 6, 8} → 5 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 5 × 8 × 7 = 280. - Case 3: Thousands digit = 6
Units digit ≠ 6 → possible units = {0, 2, 4, 8} → 4 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 4 × 8 × 7 = 224.
4: Add results
Total = 224 + 280 + 224 = 728.
Final Answer:
There are 728 such even integers.
The permutations of ${a,b,c,d,e,f,g}$ are listed in lexicographic order. Which of the following permutations are just before and just after the permutation $bacdefg$ ?
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Solution
NIMCET 2027
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and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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