Tag: combinatorics and mathematical induction

Questions Related to combinatorics and mathematical induction

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?

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

  1. $5040$

  2. $1000$

  3. $2500$

  4. $2060$

Show answer
Correct Option: A
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
$=$ $^{10}P _4=5040$.

If $^{10}P _r\,= 5040$, then find the value of $r$.

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

  1. $2$

  2. $4$

  3. $6$

  4. $8$

Show answer
Correct Option: B
Explanation:

Given, $^{10}P _r\,= 5040$
$\Rightarrow \cfrac{10!}{(10-r)!} =5040 $

$\Rightarrow (10-r)! =\cfrac{9!}{504}=\cfrac{9\times 8\times 7\times 6!}{504} = 6!$
$\therefore r=4$

If $ {^n}P _r $ $=$ 5040, then $(n, r)$ $= $

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

  1. (9,4)

  2. (10,4)

  3. (11,3)

  4. (11,4)

Show answer
Correct Option: B
Explanation:

$:^{n}P _{r}$ $=\dfrac{n!}{(n-r)!}$.
Now $5040$ $=10(9)(8)(7)$
$=\dfrac{10!}{(6)!}$
$=\dfrac{10!}{(10-4)!}$
$=:^{10}P _{4}$
Hence, $n=10$ and $r=4$

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? 

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

  1. N

  2. T

  3. E

  4. R

Show answer
Correct Option: D
Explanation:

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

If $ {^1}{^2}P _r  =$ 1320, then r $=$

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

  1. 2

  2. 3

  3. 4

  4. 5

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Correct Option: B

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

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

  1. $11!$

  2. $\dfrac{11!}{2!.4!}$

  3. $\dfrac{11!}{2!.4!.4!}$

  4. $ 2!.4!.4! $

Show answer
Correct Option: C
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!} $

If $n$ books can be arranged in a linear shelf in $5040$ different ways then the value of $n$ is

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

  1. $7$

  2. $8$

  3. $6$

  4. $9$

Show answer
Correct Option: A
Explanation:

If $n$ books are to be arranged, the no. of ways is equal to $=n!$.
$\Rightarrow$ $n!$ $=5040$
$\Rightarrow$ $n$ $=7$

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

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

  1. $100$

  2. $101$

  3. $99$

  4. $86$

Show answer
Correct Option: A
Explanation:

Given  $ ^nP _{100} = ^nP _{99} $ 


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

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


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

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

$\Rightarrow {n}-99=1$

$\Rightarrow {n}=100$

How many numbers can be formed by using all the given digits $1,2,8,9,3,5$ when repetition is not allowed

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

  1. $180$

  2. $120$

  3. $720$

  4. $1024$

Show answer
Correct Option: C
Explanation:

Number of ways of permuting $6$ different digits with no repetition $= 6! = 720$

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

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

  1. $2234$

  2. $2542$

  3. $2346$

  4. $2454$

Show answer
Correct Option: D
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 $=1680+756+18$
                             $=2456$