Combinations & Derivatives: Step-by-Step Solutions

by Rajiv Sharma 51 views

Hey guys! Today, we're diving into two fundamental concepts in mathematics: combinations and derivatives. We'll break down a couple of problems step-by-step, making sure you grasp the underlying principles. Let's get started!

Q1: Unraveling Combinations: Computing 15C3{ }^{15} C _3

Combinations are a crucial part of combinatorics, focusing on selecting items from a set where the order doesn't matter. This is different from permutations, where the order of selection is significant. Our main keyword here is combinations. In this first question, we're tackling the combination formula, specifically 15C3{ }^{15} C _3, which reads as "15 choose 3." This means we want to find out how many ways we can select 3 items from a set of 15, without considering the order. Understanding combinations is essential in various fields, from probability to computer science. Let's dive deep into how to calculate this!

The combination formula is given by:

nCr=n!r!(n−r)!{ }^{n} C _{r} = \frac{n!}{r!(n-r)!}

Where:

  • n is the total number of items.
  • r is the number of items to choose.
  • ! denotes the factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1).

In our case, n = 15 and r = 3. Plugging these values into the formula, we get:

15C3=15!3!(15−3)!=15!3!12!{ }^{15} C _3 = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!}

Now, let's expand the factorials:

15C3=15×14×13×12!3×2×1×12!{ }^{15} C _3 = \frac{15 × 14 × 13 × 12!}{3 × 2 × 1 × 12!}

Notice that 12! appears in both the numerator and the denominator, so we can cancel them out:

15C3=15×14×133×2×1{ }^{15} C _3 = \frac{15 × 14 × 13}{3 × 2 × 1}

Next, we can simplify the fraction:

15C3=15×14×136{ }^{15} C _3 = \frac{15 × 14 × 13}{6}

Divide 15 by 3 to get 5 and 14 by 2 to get 7:

15C3=5×7×13{ }^{15} C _3 = 5 × 7 × 13

Finally, multiply these numbers together:

15C3=455{ }^{15} C _3 = 455

So, 15C3{ }^{15} C _3 equals 455. This means there are 455 different ways to choose 3 items from a set of 15. The correct answer is B. 455. Isn't it cool how a seemingly complex problem boils down to simple arithmetic when you understand the formula? Remember, the key is to break it down step by step. This foundational knowledge of combinations will surely help you in various problem-solving scenarios!

Q2: Delving into Derivatives: Finding ddx(x2)\frac{d}{d x}(x^2) at M(1,0)

Moving on, let's tackle a problem from calculus: finding the derivative of a function. Derivatives are fundamental in calculus, representing the instantaneous rate of change of a function. This question asks us to find the derivative of the function f(x)=x2f(x) = x^2 and then evaluate it at the point M(1,0). The core concept here is derivatives. Derivatives have a wide range of applications, from physics (calculating velocity and acceleration) to economics (analyzing marginal cost and revenue). So, let's get our hands dirty with the power rule and understand how to solve this problem!

The derivative of a function, denoted as ddxf(x)\frac{d}{dx}f(x) or f′(x)f'(x), gives us the slope of the tangent line at any point on the function's graph. The power rule is a handy shortcut for finding the derivative of power functions, which are functions of the form xnx^n, where n is a constant. The power rule states:

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}

In our case, we have f(x)=x2f(x) = x^2. Applying the power rule, where n = 2, we get:

ddx(x2)=2x2−1=2x1=2x\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x

So, the derivative of x2x^2 is 2x2x. Now, we need to evaluate this derivative at the point M(1,0). Notice that the point M has coordinates (1,0), where x = 1 and y = 0. We're interested in the x-coordinate since we're evaluating the derivative with respect to x.

Plug x = 1 into the derivative:

2x=2(1)=22x = 2(1) = 2

Therefore, the value of the derivative ddx(x2)\frac{d}{dx}(x^2) at M(1,0) is 2. The correct answer is D. 2. See how the power rule made finding the derivative so much easier? Derivatives are a powerful tool in mathematics, and mastering them opens doors to solving many real-world problems. Keep practicing, and you'll become a derivative pro in no time!

In conclusion, we've tackled two important math problems today: computing a combination and finding a derivative. Remember, the key is to understand the underlying concepts and apply the appropriate formulas. Keep practicing, and you'll become a math whiz in no time! These skills are not just about getting the right answer; they're about building a solid foundation for more advanced topics and real-world applications. So, keep exploring, keep learning, and most importantly, keep having fun with math!