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Imagine from 2 As, 3 Bs and 3 Cs, you are to choose 4 objects. What are the number of different ways if...

(i) order does matter

(ii) order doesn't matter

I am familiar with multichoose (stars and bars) and also permutations with repetition (multinomial coefficients). However, I am still unable to use them to solve this question. From what I know, multichoose only applies if A, B, and C (the objects you can choose from) can be repeated indefinitely and not a fixed number of times. The permutation with repetition formula only applies if there are 10 objects and all 10 are shuffled.

Is there a way to use what I already know, or is there another formula or method? P.S. I am finding a way to answer all questions like the example above, not this question exclusively.

chematwork
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2 Answers2

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Generating functions offer methods broadly applicable to these and other questions. If you are not familiar with generating functions, you might take a look at this question and its answers: How can I learn about generating functions? To illustrate the use of generating functions, here are solutions to the two problems posed.

(i) If order does matter, we will use an exponential generating function to solve a more general problem. Let's say $a_r$ is the number of ways to arrange $r$ objects from the available collection of 2 As, 3 Bs and 3 Cs. We define the exponential generating function of $a_r$ by $$f(x) = \sum_{r=0}^{\infty} \frac{a_r}{r!} x^r$$ An arrangement of the letters is the "labeled product" of a sequence of 2 As, 3 Bs and 3Cs, so $$f(x) = \left( 1 + x + \frac{1}{2!}x^2 \right) \left( 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!} x^3 \right)^2 $$ Expanding this product, we have $$f(x) = 1+3 x+\frac{9 x^2}{2}+\frac{13 x^3}{3}+\frac{35 x^4}{12}+\frac{17 x^5}{12}+\frac{35 x^6}{72}+\frac{x^7}{9}+\frac{x^8}{72}$$ $f(x)$ solves the problem for general $r$. The particular case we want is $r=4$, and we have $$a_4 = 4! \cdot \frac{35}{12} = 70$$

(ii) If order doesn't matter, again we use a generating function to solve a more general problem. Let's say $b_r$ is the number of solutions in integers to $$x_1 + x_2 + x_3 = r$$ subject to $0 \le x_1 \le 2$, $0 \le x_2 \le 3$, and $0 \le x_3 \le 3$. This time we use an ordinary power series generating function. Define $$g(x) = \sum_{r=0}^{\infty} b_r x^r$$ and it's "easy to see" (with some practice) that $$g(x) = (1 + x + x^2) (1 +x +x^2 + x^3)^2$$ Expanding, we have $$g(x) = 1+3 x+6 x^2+9 x^3+10 x^4+9 x^5+6 x^6+3 x^7+x^8$$ which solves the problem for general $r$. The particular case we are interested in is $b_4 = 10$.

Although we expanded $f(x)$ and $g(x)$ completely for the purpose of illustration, texts about generating functions develop methods that often allow us to find the coefficent of a particular power of $x$ without completely expanding the generating function. (It seems these methods are more often successful in the case of ordinary power series generating functions than in the case of exponential generating functions.)

awkward
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From the four objects either there can be (3similiar 1different) or (2pairs of similar) or (2similiar 1different 1differnt)

1) 3s and 1d can be 

  S.    D 
  B.    A
  B.    C
  C.    B
  C.    A

They can be permuted in 4 ways 2)2s. 2s A. B A. C B. C They can be permuted in 6 ways 3) 1d. 1d. 2s A. B. C A. C. B B. C. A They can be permuted in 12 ways 1)If order does not matter then number of ways are {4+3+3} =10 2)If order matters then number of ways are{4×4+3×6+3×12}=70

Pran
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  • I appreciate the answer, but I am finding a way to solve questions with a higher magnitude, with more types of objects and also a higher number of objects chosen. – chematwork Oct 30 '19 at 06:27