Tag: permutations and combinations

Quizzes (1) Questions (61)

Questions Related to permutations and combinations

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ - } }\quad ^{ m }{ P } _{ { m }^{ - } }\quad (n-2)!$
  2. $^{ m }{ P } _{ { 2 }^{ - } }\quad ^{ n }{ P } _{ { n }^{ - } }\quad (n-2)!$
  3. $^{ n }{ P } _{ { 2 }^{ - } }\quad ^{ n }{ P } _{ { n }^{ - } }\quad (m-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P (distinct) = (n * (n-1)) * m! * (n-2)!. Q (alike) = (n-2)!. The ratio P/Q is n! * m! / (n-2)! which simplifies based on the specific arrangement constraints.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

In the word $ENGINEERIGNG if all $Es$ are not together and $Ns$ come together then number of permutations is

  1. $\dfrac{9!}{2!2!}-\dfrac{7!}{2!2!}$
  2. $\dfrac{9!}{3!2!}-\dfrac{7!}{2!2!}$
  3. $\dfrac{9!}{3!2!2!}-\dfrac{7!}{2!2!2!}$
  4. $\dfrac{9!}{3!2!2!}-\dfrac{7!}{2!2!}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

ENGINEERING has 11 letters: E(3), N(3), G(2), I(2), R(1). Total permutations with Ns together (treating NNN as one block) is 9! / (3! * 2! * 2!). Subtracting cases where Es are also together (EEE as one block) gives the result.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ 2^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$
  2. $^{ m }{ P } _{ 2^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$
  3. $^{ m }{ P } _{ 2^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 459936. The logic remains the same.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

If $3.^{n _{1}-n _{2}}P _{2}=^{n _{1}+n _{2}}P _{2}=90$, then the ordered $(n _{1},n _{2})$ is:

  1. $(8,2)$
  2. $(7,3)$
  3. $(16,8)$
  4. $(9,2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

3 * (n1-n2)(n1-n2-1) = (n1+n2)(n1+n2-1) = 90. Solving (n1+n2)(n1+n2-1) = 90 gives n1+n2 = 10. Solving 3 * (n1-n2)(n1-n2-1) = 90 gives (n1-n2)(n1-n2-1) = 30, so n1-n2 = 6. Adding equations: 2n1 = 16, n1 = 8. Then n2 = 2.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges. Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :

  1. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ m }{ P } _{ { m }^{ . } }(n-2)!$
  2. $^{ m }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(n-2)!$
  3. $^{ n }{ P } _{ { 2 }^{ . } }\quad ^{ n }{ P } _{ { n }^{ . } }(m-2)!$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 459936 and 459940.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

Number of ways in which these $16$ players can be divided into equal groups, such that when the best player is selected from each group, ${P} _{6}$ is one among them, is $(k)\dfrac{12!}{{4!}^{3}}$. The value of $k$ is:

  1. $36$
  2. $24$
  3. $18$
  4. $20$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This involves grouping 16 players into 4 groups of 4. The number of ways to divide them is 16! / ((4!)^4 * 4!). The condition regarding the best player involves combinations.