Tag: combinatorics and mathematical induction

Questions Related to combinatorics and mathematical induction

There are $20$ persons among whom are two brothers. The number of ways in which we can arrange them around a circle so that there is exactly one person between the two brothers, is

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

  1. $18!$

  2. $17!\times 2!$

  3. $18!\times 2!$

  4. $20!$

Show answer
Correct Option: A

The number of permutations of $n$ distinct objects taken $r$ together in which include $3$ particular things must occur together

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

  1. $^ { n-3 } C _ { r -3}(r-2)! \times 3 !$

  2. $^ { n } C _ { r - 3 } \times 3 !$

  3. $^ { n - 3 } p _ { r - 3 } \times 3 !$

  4. $P _ { 3 } \times ^ { n - 3 } P _ { r- 3 }$

Show answer
Correct Option: A
Explanation:
Number of ways of selection of (r-3) objects out of (n-3) objects
$\\=^{n-3}C _{r-3}$
When we bundle 3 things together,
$\\no. of \> objects=(r-3)+1$
$\\=(r-2)$
which can be arranged in (r-2)! ways also 3 objects
in the bundle can be arranged among themselves in 3! ways
$\\\therefore\>Permutation=^{n-3}C _{r-3}(r-2)!\>3!$

$2^n P _n$ is .equal to

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

  1. $(n + 1) (n + 2) ....... (2n)$

  2. $2^n[1.3.5 .....(2n - 1)]$

  3. $(2).(6).(10) .... (4n - 2)$

  4. $n!(2 ^nC _n )$

Show answer
Correct Option: A,B,C,D
Explanation:
$2^n{P _n}=\dfrac{2n!}{(2n-n)!}$
as ${^nP _r}=\dfrac{n!}{n-n!}$
             $=\dfrac{(2n)!}{n!}$
We can write it as,
$n^n[1.3.5.......(2n-1)]$
or
$2.6.10........(4n-2)$
also we $2^n{P _n}=(2n)(2n-2)........(n+1)$
and checking from option (d)
$n!(2{^nC _n})=\dfrac{n!(2n)!}{n!n!}$
                  $=2^n{P _n}$
$\therefore$ $(a),(b),(c),(d)$ all are correct.
Hence, all the answers are correct.

12 normal dice are thrown once. The number of ways in which each of the values 2,3,4,5 and 6 occurs exactly twice is : [1,1, 2,2, 3,3, 4,4, 5,5, 6,6 can come in any order]

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

  1. $\frac{(12)!}{6}$

  2. $\frac{(12)!}{2^{6}.6!}$

  3. $\frac{(12)!}{2^{6}}$

  4. none

Show answer
Correct Option: A

If $^{15}{P _{r - 1}}:{\,^{15}}{P _{r - 2}} = 3:4$ then $r=$

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

  1. $\dfrac{71}{4}$

  2. $14$

  3. $6$

  4. $8$

Show answer
Correct Option: A
Explanation:

Consider the given value of permutation,$^{n}{{P} _{r-1}}{{:}^{15}}{{P} _{r-2}}=3:4$

  $ \,\,\,\,\,\,\dfrac{^{15}{{P} _{r-1}}}{^{15}{{P} _{r-2}}}=\dfrac{3}{4} $

 $ \,\,\,\dfrac{\dfrac{15!}{(15-r+1)!}}{\dfrac{15!}{\left( 15-r+2 \right)!}}=\dfrac{3}{4} $

 $  $

 $ \,\,\,\,\,\,\dfrac{\left( 15-r+2 \right)!}{(15-r+1)!}=\dfrac{3}{4} $

 $ \,\,\,\,\,\dfrac{\left( 17-r \right)!}{(16-r)!}=\dfrac{3}{4} $

 $ \,\,\,\dfrac{\left( -17+r \right)!}{(-16+r)!}=\dfrac{3}{4} $

 $ \dfrac{\left( r-17 \right)!}{(r-16)!}=\dfrac{3}{4} $

 $ \dfrac{\left( r-17 \right)\left( r-16 \right)!}{(r-16)!}=\dfrac{3}{4} $

 $ r-17=\dfrac{3}{4} $

 $ r=\dfrac{71}{4} $

This is the required answer.

Let ${T _n}$ be the number of all possible triangles formed by joining vertices of an $n$-sided regular polygon. If ${T _{n + 1}} - {T _n} = 10$. 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. $5$

  3. $10$

  4. $8$

Show answer
Correct Option: B
Explanation:

Given $T _n=\ ^nC _3$

$T _{n+1}=\ ^{n+1}C _3$
therefore,
$T _{n+1}-T _n=\ ^{n+1}C _3-\ ^nC _3=10$
$\Rightarrow\ ^nC _2+\ ^nC _3-\ ^nC _3=10$        $[\because\ ^nC _r+\ ^nC _{r-1}=\ ^{n+1}C _r]$
$\Rightarrow\ ^nC _2=10$
$\Rightarrow\ ^nC _2=\ ^5C _2$
$\Rightarrow n=5$

The number of all possible different arrangements of the word $"BANANA"$ is

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

  1. $6!$

  2. $6!\times2!\times3!$

  3. $\dfrac{6!}{2!3!}$

  4. none of these

Show answer
Correct Option: C

If $\displaystyle \overset{n-r}{\underset{k=1}{\sum }}\ ^{n-k}C _r=^{x}C _y$ then-

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

  1. $x=n+1\ ;\ y=r$

  2. $x=n\ ;\ y=r+1$

  3. $x=n\ ;\ y=r$

  4. $x=n+1\ ;\ y=r+1$

Show answer
Correct Option: C

If ${}^{15}{P _{r - 1}}\,:\,{}^{15}{P _{r - 2}} = 3:4$, then $r =$

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

  1. $10$

  2. $14$

  3. $20$

  4. $15$

Show answer
Correct Option: C
Explanation:
${}^{15}{P _{r - 1}}\,:\,{}^{15}{P _{r - 2}} = 3:4$

$=>\dfrac{\dfrac{15!}{(15-(r-1))!}}{\dfrac{15!}{(15-(r-2))!}}=\dfrac{3}{4}$


$=>\dfrac{17-r}{16-r}=\dfrac{3}{4}$

$=>68-4r=48-3r$

$=>r=20$

The number of arrangements of $A _{1},A _{2},..A _{10}$ in a line so that $A _{1}$ is always above then $A _{2}$, is 

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

  1. $2\times 10!$

  2. $\dfrac{1}{2}\times10!$

  3. $^{10}P _{2}$

  4. $^{10}C _{2}$

Show answer
Correct Option: A