Tag: exponent of a prime in n!

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

The number of words that can be formed by using the letter of the word "MATHEMATICS", taken all at a time is

  1. $11!$
  2. $\dfrac{11!}{2!+2!+2!}$
  3. $\dfrac{11!}{(2!)^{3}}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The word MATHEMATICS has 11 letters, with M appearing twice, A twice, and T twice. The number of distinct permutations is 11! divided by the factorial of the counts of repeated letters, which is 11! / (2! * 2! * 2!).

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

The given relation is  $1.P(1,)+2.P(2,2)+3.P(3,3)+......+n.P(n,n)=P)(n+1,n+1)-3$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given relation has garbled notation with P)(n+1,n+1)-3 which appears to be a typo. For the standard identity sum of k×P(k,k) from k=1 to n, the right side should be P(n+1,n+1)-1, not -3. Testing with n=3: LHS = 1×1 + 2×2 + 3×3 = 14, but RHS with -3 would give P(4,4)-3 = 24-3 = 21 ≠ 14. The statement is FALSE.

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

If $^{ 56 }{ { P } _{ r+6 } }:^{ 54 }{ { P } _{ r+3 }}=30800$, then $r$ is

  1. $39$
  2. $41$
  3. $28$
  4. $43$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$^{ 56 }{ P } _{ r+6 }:^{ 54 }{ P } _{ r+3 }=30800$
$\cfrac { \cfrac { 56! }{ \left( 50-r \right) ! }  }{ \cfrac { 54! }{ \left( 51-r \right) ! }  } =30800$
$\cfrac { 56!\times \left( 51-r \right) ! }{ 54!\left( 50-r \right) ! } =30800$
$56\times 55\times \left( 51-r \right) =30800$
$\left( 51-r \right) =\cfrac { 30800 }{ 56\times 55 }$
$\left( 51-r \right) =10$
$r=41$
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

In how many ways unique can arrange the  letters  in the word "SUCCESSFUL" 

  1. $\dfrac{10!}{2!2!3!}$
  2. $\dfrac{7!}{2!2!}. ^8P _3$
  3. $^8P _3$
  4. $\dfrac{7!}{2!2!}. \dfrac{^8P _3}{3!}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
In the word 'SUCCESSFUL',
We have to arrange $10$ letters, out of which S occurs thrice, U and C occurs twice.
$\therefore$ Number of ways of arranging these letters $= \cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$

Hence the correct answer is $\cfrac{10!}{\left( 3! \right) \left( 2! \right) \left( 2! \right)}$.

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

How many $4$-letter words, with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

  1. $5040$
  2. $1000$
  3. $2500$
  4. $2060$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are $10$ letters in the word 'LOGARITHMS'.
So, the number of $4$-letter word$=$Number of arrangements of $10$ letters, taken $4$ at a time
$=$ $^{10}P _4=5040$.

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

If the last four letters of the word 'CONCENTRATION' are written in reverse order followed by next two in the reverse order and next three in the reverse order and then followed by the first four in the reverse order counting from the end which letter would be eighth in the new arrangement? 

  1. N

  2. T

  3. E

  4. R
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The new letter sequence is 
NOITARTNECNOC
The eighth letter from the end is R