Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)

In summary, the conversation discusses the use of binomial coefficients to prove the equation 3^k C(n,k) = 3^k 1^n-k C(n,k), with the hint that this is related to the coefficients of x^k in the binomial expansion of (3x + y)^n. It is noted that the coefficients of x^k in (x + y)^n are also known as binomial coefficients, and the conversation ends with a question about the coefficients of x^k in (3x + y)^n when x = y = 1.
  • #1
jenny Downer
4
0
how do i use binomial to show that 3^k C(n,k) = 3^k 1^n-k C(n,k)
 
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  • #2
Did you type that correctly? Right now you have stuff = stuff * 1, which is clearly correct...
 
  • #3
i wanted to explain a bit more
 
  • #4
It should be
n
sum 3^k C(n,k) = 2^2n
k=0

the hint is 3^k C(n,k) = 3^k 1^n-k C(n,k)
 
  • #5
The reason C(n, k) are called "binomial coefficients" is that C(n, k) is the coefficient of xk in (x+ y)n What are the coefficients of xk in (3x+ y)n? What do you get if x= y= 1?
 

Related to Using the Binomial Theorem to Show 3^k C(n,k) = 3^k 1^n-k C(n,k)

Question 1: How is the binomial theorem used to show 3^k C(n,k) = 3^k 1^n-k C(n,k)?

The binomial theorem is used to expand a binomial expression raised to a power, using the coefficients from Pascal's triangle. In this case, we can use the binomial theorem to expand (3 + 1)^n, and then substitute k for both 3 and 1, leading to the equation 3^k C(n,k) = 3^k 1^n-k C(n,k).

Question 2: What is the significance of the term 3^k C(n,k) in the equation?

The term 3^k C(n,k) represents the number of ways to choose k objects from a set of n objects, where repetition is allowed. This is also known as a combination with repetition. In the context of the equation, it is used to show that the number of combinations is equal to 3^k times the number of combinations without repetition, 1^n-k C(n,k).

Question 3: How does the equation demonstrate the relationship between combinations with repetition and without repetition?

The equation shows that the number of combinations with repetition, 3^k C(n,k), is equal to 3^k times the number of combinations without repetition, 1^n-k C(n,k). This relationship can be explained by the fact that when choosing k objects from a set of n, we can first choose k objects from the set of 3 objects (represented by 3^k), and then choose the remaining n-k objects from the set of 1 object (represented by 1^n-k), resulting in the total number of combinations.

Question 4: Can this equation be used in real-world applications?

Yes, this equation can be used in various real-world applications, such as in probability and statistics. For example, it can be used to calculate the probability of getting a specific combination of numbers when rolling multiple dice or drawing multiple cards from a deck with replacement. It can also be used in genetics to calculate the probability of inheriting certain traits from parents with a combination of dominant and recessive genes.

Question 5: Are there any limitations or assumptions to consider when using this equation?

One limitation is that this equation only applies to choosing k objects from a set of n objects with repetition allowed. It cannot be used for situations where repetition is not allowed, such as when drawing cards from a deck without replacement. Additionally, this equation assumes that all objects in the set are distinct and equally likely to be chosen. If this is not the case, the equation may not accurately represent the number of combinations.

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