Tag: combinatorics and mathematical induction

Questions Related to combinatorics and mathematical induction

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

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

  1. $27$

  2. $29$

  3. $31$

  4. $33$

Show answer
Correct Option: C
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.

If $^{56}P _{r+6} : ^{54}P _{r+3} = 30800:1$ find $r$.

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

  1. $1280$

  2. $1440$

  3. $1520$

  4. $1640$

Show answer
Correct Option: D
Explanation:

We have:
$\dfrac { 56! }{ (50-r)! } .\dfrac { (51-r)! }{ 54! } =30800\ =>56.55.(51-r)=30800\ =>51-r=10=>r=41$

Therefore, $ _{ 2 }^{ r }{ P }= _{ 2 }^{ 41 }{ P }=41.40=1640$
Hence, (D) is correct.

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

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

  1. $\dfrac { 7 } { 12 }$

  2. $\dfrac { 5 } { 12 }$

  3. $\dfrac { 1 } { 3 }$

  4. $\dfrac { 1 } { 4 }$

Show answer
Correct Option: C
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.

Which of the following is true ?

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

  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

Show answer
Correct Option: A
Explanation:

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

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

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

$\quad = \space ^nP _r$

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

If $\displaystyle \frac{^{n}P _{r-1}}{a}=\frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$,then which of the following holds good 

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

  1. $c^{2}=a(b+c)$

  2. $a^{2}=c(a+b)$

  3. $b^{2}=a(b+c)$

  4. $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

Show answer
Correct Option: C
Explanation:

$\displaystyle \frac{^{n}p _{r-1}}{a}=\frac{^{n}P _{r}}{b}$

$\displaystyle \Rightarrow n-r=\frac{b}{a}-1$

and $\displaystyle \frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$

$\displaystyle \Rightarrow n-r=\frac{c}{b}$
On dividing (i) and (ii) we get
$b^{2}=a(b+c)$