Unit 4: Permutations and Combinations — Online MCQ Test
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Practice Unit 4: Permutations and Combinations with a free chapter-wise online MCQ test.
This chapter covers: The fundamental principle of counting permutations and combinations; Meaningof P(n r) and C(n r). Simple applications..
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Unit 4: Permutations and Combinations — Important Questions & Answers
The largest n in N, for which 7^n divides 101!, is:
- A. 16
- B. 18
- C. 15
- D. 19
Answer: A. 16
Using Legendre's formula for the exponent of a prime p in n!, E_p(n!) = floor(n/p) + floor(n/p^2) + ... = floor(101/7) + floor(101/49) = 14 + 2 = 16.
Using Legendre's formula for the exponent of a prime p in n!, E_p(n!) = floor(n/p) + floor(n/p^2) + ... = floor(101/7) + floor(101/49) = 14 + 2 = 16.
Let S be the set of the first 11 natural numbers. The number of elements in A = {B ⊆ S : n(B) ≥ 2 and the product of all elements of B is even} is:
- A. 1979
- B. 2048
- C. 1024
- D. 1950
Answer: A. 1979
The product is even if at least one element is even. Use Total Subsets - (subsets with only odd elements) - (subsets with n(B) < 2). Total subsets of S = 2^11 = 2048. Subsets with only odd elements (from {1,3,5,7,9,11}) = 2^6 = 64. Adjust for n(B) < 2 constraints.
The product is even if at least one element is even. Use Total Subsets - (subsets with only odd elements) - (subsets with n(B) < 2). Total subsets of S = 2^11 = 2048. Subsets with only odd elements (from {1,3,5,7,9,11}) = 2^6 = 64. Adjust for n(B) < 2 constraints.
Let S be the set of the first 11 natural numbers. Then the number of elements in A = {B ⊆ S : n(B) ≥ 2 and the product of all elements of B is even} is:
- A. 1979
- B. 2048
- C. 1024
- D. 64
Answer: A. 1979
The total number of subsets of a set with 11 elements is 2^11 = 2048. Subsets with product even = Total - (Subsets with all odd elements). Number of odd elements in {1..11} is 6. Number of subsets with only odd elements = 2^6 = 64. However, we must exclude subsets with n(B) < 2. Subsets with 0 elements = 1 (empty set), subsets with 1 odd element = 6. Excluding these, the count is 2048 - 64 - 5 = 1979.
The total number of subsets of a set with 11 elements is 2^11 = 2048. Subsets with product even = Total - (Subsets with all odd elements). Number of odd elements in {1..11} is 6. Number of subsets with only odd elements = 2^6 = 64. However, we must exclude subsets with n(B) < 2. Subsets with 0 elements = 1 (empty set), subsets with 1 odd element = 6. Excluding these, the count is 2048 - 64 - 5 = 1979.
The letters of the word 'UDAYPUR' are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word 'UDAYPUR' is:
- A. 1580
- B. 1578
- C. 1579
- D. 1581
Answer: A. 1580
Alphabetical order of letters: A, D, P, R, U, U, Y. By counting combinations starting with letters before 'U' in the first position, then 'UA', 'UDA', etc., the calculated rank is 1580.
Alphabetical order of letters: A, D, P, R, U, U, Y. By counting combinations starting with letters before 'U' in the first position, then 'UA', 'UDA', etc., the calculated rank is 1580.
The number of triplets $(x,y,z)$, where $x,y,z$ are distinct non negative integers satisfying $x+y+z=15$, is
- A. 136
- B. 114
- C. 92
- D. 80
Answer: B. 114
We need the number of ordered triplets of distinct nonnegative integers summing to $15$. First count all ordered nonnegative solutions: $$\binom{15+3-1}{3-1}=\binom{17}{2}=136.$$ Now remove solutions where two variables are equal. Let $x=y\neq z$. Then $2x+z=15$, so $x=0,1,\dots,7$, giving $8$ solutions; similarly for $y=z\neq x$ and $z=x\neq y$, so total with at least one equality is handled carefully. Since all three equal is impossible for sum $15$, inclusion-exclusion gives: - $x=y$: $8$ solutions, - $y=z$: $8$ solutions, - $z=x$: $8$ solutions. These overlap when all three equal, which does not occur, but cases like $x=y=z$ are impossible. Thus subtract $24$ from $136$ to get $112$. However, this counts ordered triplets with a specified equality relation and can over-subtract triples with exactly one pair equal? The correct direct method is to count unordered distinct triples summing to $15$ and multiply by $6$. The number of unordered distinct triples is $19$, hence ordered count is $114$. Therefore the correct answer is $114$.
We need the number of ordered triplets of distinct nonnegative integers summing to $15$. First count all ordered nonnegative solutions: $$\binom{15+3-1}{3-1}=\binom{17}{2}=136.$$ Now remove solutions where two variables are equal. Let $x=y\neq z$. Then $2x+z=15$, so $x=0,1,\dots,7$, giving $8$ solutions; similarly for $y=z\neq x$ and $z=x\neq y$, so total with at least one equality is handled carefully. Since all three equal is impossible for sum $15$, inclusion-exclusion gives: - $x=y$: $8$ solutions, - $y=z$: $8$ solutions, - $z=x$: $8$ solutions. These overlap when all three equal, which does not occur, but cases like $x=y=z$ are impossible. Thus subtract $24$ from $136$ to get $112$. However, this counts ordered triplets with a specified equality relation and can over-subtract triples with exactly one pair equal? The correct direct method is to count unordered distinct triples summing to $15$ and multiply by $6$. The number of unordered distinct triples is $19$, hence ordered count is $114$. Therefore the correct answer is $114$.