Solve the following mathematical problems:

Given, serial numbers are to be made using 2 letters followed by 3 numbers.

For the letters:

There are 8 distinct letters (first 8 letters of the alphabet: A, B, C, D, E, F, G, H).

We need to choose and arrange 2 letters without repetition.

This is a permutation of 8 items taken 2 at a time, denoted as \( P(8, 2) \).

The formula for permutation is \( P(n, k) = \frac{n!}{(n-k)!} \).

Number of ways to arrange letters = \( P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 8 \times 7 = 56 \)

For the numbers:

There are 10 distinct digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).

We need to choose and arrange 3 numbers without repetition.

This is a permutation of 10 items taken 3 at a time, denoted as \( P(10, 3) \).

Number of ways to arrange numbers = \( P(10, 3) = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \times 9 \times 8 = 720 \)

To find the total number of possible serial numbers, we multiply the number of ways to arrange the letters by the number of ways to arrange the numbers (Fundamental Principle of Counting).

Total number of serial numbers = (Number of ways to arrange letters) \(\times\) (Number of ways to arrange numbers)

Total number of serial numbers = \( 56 \times 720 \)

Total number of serial numbers = \( 40320 \)

Therefore, 40320 serial numbers are possible.

Step 1: Understand the problem and identify the given values.

Let the height of the tower be AB. Given, \(AB = h = 150\sqrt{3}\) m.

Let the two objects be at points C and D, located on the same side of the base of the tower B.

The angles of depression from the top of the tower A are \(60^\circ\) and \(30^\circ\).

As the angle of depression is larger for the closer object, point C is closer to the tower than point D.

So, the angle of depression to C is \(60^\circ\), which implies \(\angle ACB = 60^\circ\).

The angle of depression to D is \(30^\circ\), which implies \(\angle ADB = 30^\circ\).

We need to find the distance between the two objects, i.e., CD.

Step 2: Calculate the distance of the closer object (C) from the base of the tower (B).

Consider the right-angled triangle \(\triangle ABC\).

We know that \(\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}\).

Here, \(\tan(\angle ACB) = \frac{AB}{BC}\).

\(\tan(60^\circ) = \frac{150\sqrt{3}}{BC}\)

\(\sqrt{3} = \frac{150\sqrt{3}}{BC}\)

\(BC = \frac{150\sqrt{3}}{\sqrt{3}}\)

\(BC = 150\) m

Step 3: Calculate the distance of the farther object (D) from the base of the tower (B).

Consider the right-angled triangle \(\triangle ABD\).

Here, \(\tan(\angle ADB) = \frac{AB}{BD}\).

\(\tan(30^\circ) = \frac{150\sqrt{3}}{BD}\)

\(\frac{1}{\sqrt{3}} = \frac{150\sqrt{3}}{BD}\)

\(BD = 150\sqrt{3} \times \sqrt{3}\)

\(BD = 150 \times 3\)

\(BD = 450\) m

Step 4: Determine the distance between the two objects.

The distance between the two objects (CD) is the difference between BD and BC.

\(CD = BD - BC\)

\(CD = 450 - 150\)

\(CD = 300\) m

Therefore, the distance between the two objects is \(300\) meters.

Let the first term of the geometric series be \(a\) and the common ratio be \(r\).

The \(n\)-th term of a geometric series is given by the formula:

\[T_n = a r^{n-1}\]

According to the given problem:

4^th term, \(T_4 = a r^{4-1} = a r^3 = \frac{\sqrt{2}}{3}\)   —  (1)

10^th term, \(T_{10} = a r^{10-1} = a r^9 = \frac{8\sqrt{2}}{81}\)   —  (2)

Dividing equation (2) by equation (1):

\[\frac{a r^9}{a r^3} = \frac{\frac{8\sqrt{2}}{81}}{\frac{\sqrt{2}}{3}}\]

\[r^6 = \frac{8\sqrt{2}}{81} \times \frac{3}{\sqrt{2}}\]

\[r^6 = \frac{8}{27}\]

Taking the square root on both sides to find \(r^3\):

\[r^3 = \sqrt{\frac{8}{27}} = \frac{2\sqrt{2}}{3\sqrt{3}}\]

Now, the 7^th term of the series is:

\[T_7 = a r^{7-1} = a r^6 = (a r^3) \cdot r^3\]

Substituting the values of \(a r^3\) and \(r^3\):

\[T_7 = \left(\frac{\sqrt{2}}{3}\right) \times \left(\frac{2\sqrt{2}}{3\sqrt{3}}\right)\]

\[T_7 = \frac{2 \times 2}{9\sqrt{3}} = \frac{4}{9\sqrt{3}}\]

Rationalizing the denominator by multiplying both the numerator and denominator by \(\sqrt{3}\):

\[T_7 = \frac{4\sqrt{3}}{27}\]

Answer: \(\frac{4}{9\sqrt{3}}\) (or \(\frac{4\sqrt{3}}{27}\))

To find the number nearest to 43583 that is divisible by 25, 50, and 75, we first need to calculate the Least Common Multiple (L.C.M.) of 25, 50, and 75.

Step 1: Find the L.C.M. of 25, 50, and 75

Prime factorizations of the numbers:

\(25 = 5^2\)

\(50 = 2 \times 5^2\)

\(75 = 3 \times 5^2\)

\(\text{L.C.M.}(25, 50, 75) = 2 \times 3 \times 5^2 = 150\)

Step 2: Divide 43583 by the L.C.M. (150) to find the remainder

\(43583 \div 150 = 290\) with a remainder of \(83\)

Step 3: Compare the nearest multiples of 150

1. The preceding multiple of 150 (smaller than 43583):

\(43583 - 83 = 43500\)

Difference from 43583: \(83\)

2. The succeeding multiple of 150 (larger than 43583):

\(43583 + (150 - 83) = 43583 + 67 = 43650\)

Difference from 43583: \(67\)

Conclusion:

Since \(67 < 83\), the number 43650 is closer to 43583 than 43500.

Therefore, the number nearest to 43583 which is divisible by each of 25, 50, and 75 is 43650.

Given the following information:

Principal debt (PV) = Tk. 1000

Annual interest rate = 16%

Compounded quarterly

Number of quarterly payments (n) = 20 (over 5 years, with 4 quarters/year, so 5 * 4 = 20 payments)

First, calculate the quarterly interest rate (i):

\(i = \frac{\text{Annual Interest Rate}}{\text{Number of compounding periods per year}}\)

\(i = \frac{16\%}{4} = 4\% = 0.04\)

The formula for calculating the payment (PMT) of an amortized loan (present value of an ordinary annuity) is:

\(PMT = PV \times \frac{i}{1 - (1+i)^{-n}}\)

Now, substitute the known values into the formula:

\(PMT = 1000 \times \frac{0.04}{1 - (1+0.04)^{-20}}\)

\(PMT = 1000 \times \frac{0.04}{1 - (1.04)^{-20}}\)

Calculate \((1.04)^{-20}\):

\((1.04)^{-20} \approx 0.45638693\)

Substitute this value back into the equation:

\(PMT = 1000 \times \frac{0.04}{1 - 0.45638693}\)

\(PMT = 1000 \times \frac{0.04}{0.54361307}\)

Calculate the fraction:

\(\frac{0.04}{0.54361307} \approx 0.07358703\)

Finally, calculate the payment amount (PMT):

\(PMT = 1000 \times 0.07358703\)

\(PMT \approx 73.58703\)

Rounding to two decimal places, the amount of each quarterly payment will be Tk. 73.59.