Question

Given

• N = 5 // n is the number of operations

• A = [1, 5, 4, 2, 3, 6]

Approach

• There are 759375 possible sets of 5 operations that can be applied on array A out of which only 2848 sets of operations yield a sorted array.

can anyone please solve this permutation probability​

Answer 1

Answer:

$\large\underline{\sf{Solution-}}$

Given that,

A man invested Rs. 20000 at 10% per annum at simple interest.

Another amount at 5% per annum at simple interest.

At the end of the year he got 7% interest on the entire investment.

Let assume that

The amount invested at the rate of 5 % per annum be Rs x

Case :- 1

Principal, P = Rs 20000

Rate of interest, r = 10 % per annum

Time, n = 1 year

We know,

Simple interest (SI) received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

$\boxed{ \rm{ \:SI \: = \: \frac{P \times r \times n}{100} \: }}$

So, on substituting the values, we get

$\rm \: SI_1 \: = \: \dfrac{20000 \times 10 \times 1}{100}$

$\rm\implies \:\boxed{ \rm{ \:SI_1 \: = \: Rs \: 2000 \: \: }} - - - (1)$

Case :- 2

Principal, P = Rs x

Rate of interest, r = 5 % per annum

Time, n = 1 year

So,

$\rm \: SI_2 = \dfrac{x \times 5 \times 1}{100}$

$\rm\implies \:\boxed{ \rm{ \:SI_2 \: = \: \frac{5x}{100} \: \: }} - - - (2)$

Case :- 3

Principal, P = Rs (20000 + x)

Rate of interest, r = 7 % per annum

Time, n = 1 year

So,

$\rm \: SI_3 \: = \: \dfrac{(20000 + x) \times 7 \times 1}{100}$

$\rm\implies \:\boxed{ \rm{ \:SI_3 \: = \: \frac{140000 + 7x}{100} \: }} - - - (3)$

Now, According to statement

$\rm \: SI_3 = SI_1 + SI_2$

On substituting the values from equation (1), (2) and (3), we get

$\rm \: \dfrac{140000 + 7x}{100} = 2000 + \dfrac{5x}{100}$

$\rm \: \dfrac{140000 + 7x}{100} = \dfrac{200000 + 5x}{100}$

$\rm \: 140000 + 7x = 200000 + 5x$

$\rm \: 7x - 5x = 200000 - 140000$

$\rm \: 2x = 60000$

$\rm\implies \:x = 30000$

So,

$\rm \:Total\:investment \: = \: 20000 +x$

$\rm \:Total\:investment \: = \: 20000 +30000$

$\rm\implies \:\boxed{ \bf{ \:Total\:investment \: = \: Rs \: 50000 \: }}$

$\rm\implies \:$Option (d) is correct

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{P = \dfrac{SI \times 100}{r \times n} }\\ \\ \bigstar \: \bf{r = \dfrac{SI \times 100}{P \times n} }\\ \\ \bigstar \: \bf{n = \dfrac{SI \times 100}{P \times r} }\\ \\ \bigstar \: \bf{Amount = P\bigg(\dfrac{100 + rn}{100} \bigg) }\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

• N = 5 // n is the number of operations

So ,n^5-n= n(n^4-1)=n*(n^2+1)*(n+1)(n-1)

Immediately we can see that n,n-1,and n-2 are consecutive numbers,and knowing that theres always a multiple of 2 amongst two consecutive numbers and a multiple of three amongst three consecutive numbers ,we get that our given expression is divisible by (2*3)=6.

We only need to show that the expression is divisible by 5 now. So let's break it into 5cases.

1)n=5k for some integer k . this is trivial

2)n=5k+1,this means n-1 is a factor of 5

3)n=5k+2 ,this means n^2+1= 25k^2+20k+4+1

=5(5k^2+4k+1) which is obviously a multiple of 5

4)n=5k+3, similar to 3) n^2+1 equals

25k^2+30k+10 ,which is a multiple of 5

5)n=5k+4, again n+1 becomes a multiple of 5

Thus we have proven that for all values of n the expression will always be divisible by 5,and knowing that the expression is also always divisible by 6, we an sat that its is always divisible by (5*6)=30

A = [1, 5, 4, 2, 3, 6]

it's Answer is in the attachment.

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