Tag: combinatorics and mathematical induction

Quizzes (3) Questions (65)

Questions Related to combinatorics and mathematical induction

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

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

How many words, with meaning or without meaning, can be formed by using the letters of the word $'MISSISSIPPI'$

  1. $11!$
  2. $\dfrac{11!}{2!.4!}$
  3. $\dfrac{11!}{2!.4!.4!}$
  4. $ 2!.4!.4! $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of letters in the word $=  11$. Among them $ 2P, 4S, 4I$ are there.
Total number of words that can be formed $= \dfrac{11!}{2!.4!.4!} $

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

If $ ^nP _{100} = ^nP _{99} $, then $n$ is equal to

  1. $100$
  2. $101$
  3. $99$
  4. $86$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given  $ ^nP _{100} = ^nP _{99} $ 


Formula: $^np _r=\dfrac{n!}{(n-r)!}$

$\Rightarrow \displaystyle \dfrac{n!}{(n-100)!}=\frac{n!}{(n-99)!}$


$\Rightarrow (n-99)!=(n-100)!$

$\Rightarrow (n-99)(n-100)!=(n-100)!$

$\Rightarrow {n}-99=1$

$\Rightarrow {n}=100$

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

How many words can be formed by taking $4$ letters at a time of the letters of word $MATHEMATICS$ 

  1. $2234$
  2. $2542$
  3. $2346$
  4. $2454$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given words $MATHEMATICS$
for our simplicity we can write it as
$M\ A\ T\ H\ E\ I\ C\ S\ $
$M\ A\ T\ $
a) words of type $ABCD$
(All four letters are different)
(4 letters are chosen from $MATHEICS$ and are arranged)
$={ 8 } _{ { c } _{ 4 } }\times 4!=1680$

b) words of type $AABC$
(2 are alike, 2 are different)
(1 pair of letters is selected from M, A, T and 2 letters
are chosen from the remaining 7 letters and are arranged)
$={3} _{{c} _{1}} \times {7} _{{c} _{2}}\times \dfrac{4!}{2!}=756$

c) words of type $AABB$
(2 are alike of 1 kind, 2 are alike of another kind)
(2 pair of letter are chosen from M, A, T and arranged) 
$={3} _{c _2} \times \dfrac{4!}{2!2!}=18$
Total no. of word $=1680+756+18$
                             $=2456$