Tag: permutations and combinations

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Questions Related to permutations and combinations

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

In an examination hall, there are four rows of chairs. Each row has $8$ chairs one behind the other. There are two classes sitting for the examination with $16$ students in each class. It is desired that in each row all students belong to the same class and that no two adjacent rows are allotted to the same class. In how many ways can these $32$ students be seated?

  1. $2 \times 16! \times 16!$
  2. $4 \times 16! \times 16!$
  3. $2 \times 8! \times 16!$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since there are 4 rows, let us label the rows as row 1,2,3,4.

each row has 8 chairs. since all the students of the same class sit in the same row. and no adjacent row is alloted to the same class.

therefore one class can be alloted either in 1 and 3 rows or 2 and 4 rows. therefore there are 2 ways to allot the rows to the class.

now 16 students of this class can be arranged in 16 seats, the number of ways to arrange 16 students in 16 seats=16!

similarly 16 students of other class can be arranged in 16! ways.

therefore total number of ways=2$\times$16!$\times$16! ways

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

'$X$' completes a job in $2$ days and '$Y$' completes it in $3$ days and '$Z$' takes $4$ days to complete it. If they work together and get Rs. $3,900$ for the job, then how much amount does '$Y$' get?

  1. Rs. $1,800$
  2. Rs. $ 1,200$
  3. Rs. $ 900$
  4. Rs. $ 800$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$X$ do job in $2$ days, $Y$ completes it in $3$ days, $Z$ takes $4$ days.
If $X, Y, Z$ together can do in $1$ day, then 
$= \dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} = \dfrac {13}{12}$ of work
Therefore, the whole work is done in $\dfrac {12}{13}$ of a day.
Daily wages of $Y = \dfrac {1}{3}\times $ Rs. $ 3,900 =$ Rs. $ 1,300$
$\therefore$ Amount of $Y = \dfrac {12}{13} \times$ Rs. $ 1,300 =$ Rs. $1,200$

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

Find the number of words, with meaning or without meaning, that can be formed by arranging the letters of the word $'EIGHT'$ in all possible ways 

  1. $180$
  2. $120$
  3. $24$
  4. $720$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Number of ways of permuting $5$ distinct elements = $5!$
$\therefore$ The answer $=5! = 5\times4\times 3\times2 \times1=120$
Hence option B

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

Find the number of ways in which the letters of the word $'AEROPLANE'$ can be arranged such that the vowels are always together.

  1. $\dfrac{5!}{2!}$
  2. $\dfrac{5!}{2!^2}$
  3. $\dfrac{5!^2}{2!^2}$
  4. $\dfrac{9!}{2!^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$'AEROPLANE'$ Total number of letters $=9$
Vowels - $ 2A,2E,1O$
Considering all vowels as a single letter, we are left with $5$ letters
They can be arranged in $5!$ ways 
Now vowels can be arranged among themselves in $\dfrac{5!}{2!.2!}$
So required number of ways $=\dfrac{5!^2}{2!^2} = 3600 $

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

In how many ways can you partition $6$ into ordered summands? (For example, $3$ can be partitioned in $3$ ways as : $1 + 2, \,2 + 1, \,1 + 1 + 1$)

  1. $27$
  2. $29$
  3. $31$
  4. $33$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$1+1+1+1+1+1$      $1$ ways

$2+1+1+1+1$             $5$ ways
$3+1+1+1$                    $4$ ways
$4+1+1$                           $3$ ways
$5+1$                                  $2$ ways
$2+3+1$                           $6$ ways
$2+4$                                  $2$ ways
$3+3$                                  $1$ ways
$2+2+2$                           $1$ ways
$2+2+1+1$                    $6$ ways
Total number of ways of partitioning $6=1+5+4+3+2+6+2+1+1+6=31$.
Hence, option C is correct.

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

A bag contains  $4$ red,  $3$ black, and  $2$ white balls. If  $2$  balls are selected at random, the probability of selecting atleast one white ball is

  1. $\dfrac { 7 } { 12 }$
  2. $\dfrac { 5 } { 12 }$
  3. $\dfrac { 1 } { 3 }$
  4. $\dfrac { 1 } { 4 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Bag contains $4$ red, $3$ black and, $2$ white balls
Two balls are selected at random
The total no. of ways of doing that is ${ 10 } _{ { C } _{ 2 } }=\dfrac { 10! }{ 8!\times 2! } =\dfrac { 10\times 9\times 8! }{ 8!\times 2! } =\dfrac { 10\times 9 }{ 2 } =45$
Now in the selection we need to ensure that at least on white ball is selected.
Case $1$ :  $1$ white ball $+$ $1$ ball of any other color
This can be done in ${ 2 } _{ { C } _{ 1 } }\times { 7 } _{ { C } _{ 1 } }=2\times 7=14$ ways
Case $2$ :  $2$ white ball
This can be done in ${ 2 } _{ { C } _{ 2 } }=1$ way
$\therefore$   probability of selecting atleast one white ball is $=\dfrac { 14+1 }{ 45 } =\dfrac { 15 }{ 45 } =\dfrac { 1 }{ 3 } $
Answer : Option C.
Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

Which of the following is true ?

  1. $^nP _r = ^{n-1}P _r + r\times ^{n-1}P _{r-1}$
  2. $^nP _r = ^{n-1}P _{r-1} + r\times ^{n-1}P _{r-1}$
  3. $^nC _r = ^{n-1}C _{r-1} + r \times^{n-1}C _{r-1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\mbox{R.H.S. =}  ^{n-1}P _r + r ^{n-1}P _{r-1}$
$\quad = \displaystyle\frac{(n-1)!}{(n-1-r)!} + r\displaystyle\frac{(n-1)!}{(n-1-r+1)!} = \displaystyle\frac{(n-1)!}{(n-r-1)!} + \displaystyle\frac{r(n-1)!}{(n-r)!}$

$\quad = \displaystyle\frac{(n-r)(n-1)!}{(n-r)(n-1-r)!} + \displaystyle\frac{r(n-1)!}{(n-1)!} = \displaystyle\frac{(n-r)(n-1)!}{(n-r)!} + \displaystyle\frac{r(n-1)!}{(n-1)!} \quad [\therefore \space\alpha(\alpha - 1)! = \alpha!, \alpha\in N]$

$\quad = \displaystyle\frac{(n-1)!}{(n-r)!}[n - r + r] = \displaystyle\frac{n(n-1)!}{(n-1)!} = \displaystyle\frac{n!}{(n-r)!}$

$\quad = \space ^nP _r$

$\quad = \space \mbox{L.H.S}$