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Original Question:

For the strings of length 10 in how many have:

  • a) four O's, three 1 's, and three 2'?
  • b) at least eight l's ?
  • c) weight 4 ?

Textbook Solution:

a) 10!/4!3!3! (arrangements with repeated symbols)

b)$10\choose8$$2^2$ + $10\choose9$$2^1$ + $10\choose10$$2^0$

c)$10\choose4$(four 1's, six 0's) + $10\choose2$$8\choose1$ (two 1's, one 2, seven 0's) + $10\choose2$(two 2's, eight 0's)

Textbook's Background Information: (Hint)

  • Assume the string is the length 10, and it is made of 0's, 1's, and 2's.
  • "In general, if n is any positive integer, then by the rule of product there are $3^n$ strings of length n for the alphabet 0, 1, 2".
  • "If x = x 1 x2 x3 · · · Xn is one of these strings, we define the weight ofx, denoted wt(x), bywt(x) = x 1 + x2 + x3 + · · · + Xn. For example, wt(l2) = 3 and wt(22) = 4 for the case where n = 2; wt(101) = 2, wt(210) = 3, and wt(222) = 6 for n = 3".

Personal Logical Question:

I have problem understanding the solution and logic for part C. My guess would be weight of 4 is referring to four 1's or two 1's and one 2's or two 2's. Am I right ?

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    I provided my textbook's background information, so those helpful people who want to clarify part c for me have an easier time understanding what is the question asking me. There is no useless information. – 2017-02-17

1 Answers 1

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The weight is just the sum of the entries in the string. It appears that the strings are only composed of $0,1,2$, though you don't say that. In that case, a weight of four can then be achieved by four $1$s, a $2$ and two $1$s, or two $2$s and the rest zeros. The text is counting the number of ways to positions those in a string of $10$ for each case.

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    Yes I agree with you. I just didn't understand what you mean here: "The text is counting the number of ways to positions those in a string of 10 for each case". – 2017-02-17
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    For one two and two ones, there also seven zeros to make a string of length ten. The ${10 \choose 2}$ is the number of ways to place the ones in the string and the $8 \choose 1$ is the number of ways to place the two in what is left. It gives the same result as $\frac {10!}{2!1!7!}$ following the logic in a – 2017-02-17
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    How would we express the final answer, $10\choose4$(four 1's, six 0's) + $10\choose2$$8\choose1$ (two 1's, one 2, seven 0's) + $10\choose2$(two 2's, eight 0's), in the form of arrangements with repeated symbols like this one --> 10!/2!1!7! ? – 2017-02-17
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    Just use the formula for [combinations](https://en.wikipedia.org/wiki/Combination), so ${10 \choose 4}=\frac {10!}{4!6!}$ You will still have a sum of three terms of that sort. – 2017-02-17
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    Yeah right, I just made a minor mistake, but it gives the right answer. – 2017-02-17