Exponent of a prime in n! - class-XI

The word MATHEMATICS has 11 letters, with M appearing twice, A twice, and T twice. The number of distinct permutations is 11! divided by the factorial of the counts of repeated letters, which is 11! / (2! * 2! * 2!).

The given relation has garbled notation with P)(n+1,n+1)-3 which appears to be a typo. For the standard identity sum of k×P(k,k) from k=1 to n, the right side should be P(n+1,n+1)-1, not -3. Testing with n=3: LHS = 1×1 + 2×2 + 3×3 = 14, but RHS with -3 would give P(4,4)-3 = 24-3 = 21 ≠ 14. The statement is FALSE.

The sum 11! + 22! + ... + n*n! equals (n+1)! - 1. The relation given is (n+1)! - 3, which is incorrect.

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
$=$ $^{10}P _4=5040$.

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

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

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

The notation is garbled but appears to mean 12Pr = 1320. 12P3 = 12×11×10 = 1320. Therefore r=3. For r=2: 12P2 = 12×11 = 132. For r=4: 12P4 = 12×11×10×9 = 11880. Only r=3 gives 1320, making B the correct answer.

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

Given  $ ^nP _{100} = ^nP _{99} $ 

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

To have exactly one person between the two specific brothers in a circle of 20 people, we can treat the two brothers and the person between them as a single fixed block or use combinatorial placement. Specifically, choose the person between them in 18 ways, place the two brothers in 2! ways relative to that person, and arrange the remaining 17 entities around the circle in (18 - 1)! = 17! ways, leading to 18 * 2! * 17! = 18! * 2!.

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.

Given $T _n=\ ^nC _3$

The word BANANA has 6 letters: B(1), A(3), N(2). The number of arrangements is 6! / (1! * 3! * 2!).

Using the hockey stick identity (or repeated application of Pascal's identity C(n,r) + C(n,r+1) = C(n+1,r+1)), the sum of combinations simplifies to C(n, r+1). Therefore, x = n and y = r + 1.

By symmetry, for any linear arrangement of 10 distinct elements, the condition that A1 appears before (or above) A2 is equally likely as A2 appearing before A1. Thus, exactly half of the total 10! arrangements satisfy the given condition.

  Given $^{2n+1}P _{n-1}: ^{2n-1}P _n = 3 : 5$ which can be further simplified as
$ \displaystyle \frac {\frac { (2n+1)! }{ (n+2)! }  }{ \frac { (2n-1)! }{ (n-1)! }  }= \frac { 3 }{ 5 } $
$\Rightarrow \displaystyle \frac {(2n+1)(2n)}{(n+2)(n+1)(n)} = \displaystyle\frac{3}{5} $
$\Rightarrow $ $3{n}^{2}-11n-4=0$ .
On solving this quadratic equation we get roots as  $ 4 $ and $ -\displaystyle \frac{1}{3} $.
Since $n$ is an integer,$n=4$.

The digits are 1, 2, 2, 3, 3, 3, 4. Total digits = 7. Counts: 1(1), 2(2), 3(3), 4(1). Permutations = 7! / (1! * 2! * 3! * 1!) = 5040 / (2 * 6) = 5040 / 12 = 420.

P (distinct) = (n * (n-1)) * m! * (n-2)!. Q (alike) = (n-2)!. The ratio P/Q is n! * m! / (n-2)! which simplifies based on the specific arrangement constraints.

Exponent of $4$ is $80!$ is

Using the relation P(n, r) = n * P(n-1, r-1), we expand P(n, 5) as n * P(n-1, 4). Equating this to 9 * P(n-1, 4) gives n * P(n-1, 4) = 9 * P(n-1, 4), which implies n = 9 since P(n-1, 4) is non-zero.

ENGINEERING has 11 letters: E(3), N(3), G(2), I(2), R(1). Total permutations with Ns together (treating NNN as one block) is 9! / (3! * 2! * 2!). Subtracting cases where Es are also together (EEE as one block) gives the result.

This is a duplicate of 459936. The logic remains the same.

3 * (n1-n2)(n1-n2-1) = (n1+n2)(n1+n2-1) = 90. Solving (n1+n2)(n1+n2-1) = 90 gives n1+n2 = 10. Solving 3 * (n1-n2)(n1-n2-1) = 90 gives (n1-n2)(n1-n2-1) = 30, so n1-n2 = 6. Adding equations: 2n1 = 16, n1 = 8. Then n2 = 2.

This is a duplicate of 459936 and 459940.

Expanding the ratio (2n+1)! / (n+2)! divided by (2n-1)! / (n-1)! = 3/5. This simplifies to (2n+1)(2n) / (n+2)(n+1)(n) = 3/5. Solving for n gives n=4.

This involves grouping 16 players into 4 groups of 4. The number of ways to divide them is 16! / ((4!)^4 * 4!). The condition regarding the best player involves combinations.

A one-to-one function from a set of 3 elements to a set of 5 elements is equivalent to choosing 3 elements from 5 and arranging them, which is 5P3.

For a garland (necklace where clockwise and counterclockwise arrangements are distinct or identical depending on convention, but standard formula applies), the number of ways to arrange 8 objects with 4 specific objects always together is ((8 - 4 + 1 - 1)! * 4!) / 2 = (4! * 4!) / 2 = (24 * 24) / 2 = 288.

Applying the given condition, we get
$\dfrac{(n+5)!}{4!}=\dfrac{11(n-1)}{2}\dfrac{(n+3)!}{3!}$
$\dfrac{(n+4)(n+5)}{4}=\dfrac{11(n-1)}{2}$
$(n+4)(n+5)=22(n-1)$
$n^{2}+9n+20=22n-22$
$n^{2}-13n+42=0$
$(n-7)(n-6)=0$
$n=7$ $n=6$

$^nP _r = \dfrac { n! }{ (n-r)! } $

So, $ ^nP _n = \dfrac { n! }{ (n-n)! } =\dfrac { n! }{ 0! }  = n ! $ (Since $ 0 ! = 1 )$

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

$ { { n } _{ P } } _{ r } = \frac { n! }{ (n-r)! } $

So, $ { { 5 } _{ P } } _{ 4 } = \frac { 5! }{ (5-4)! } =\frac { 5! }{ 1! }  = 5 ! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $ (Since $ 1 ! = 1 )$

Number of letters in the word $= 6$, $E$ is repeated twice
So, total number of ways of arranging $=\dfrac{6!}{2!} $

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

$'AEROPLANE'$ 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 $

$1+1+1+1+1+1$      $1$ ways

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.

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

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

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

$\quad = \space ^nP _r$

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

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

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

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