A list of 10 unique words is randomized twice. Both times, the 10 words are written down from left to right. What is the probability that none of the first three words in the first randomized trial match the first three words in the second randomized trial?. Answer: 7/24
I didn’t like the explanation given. Is there a way to solve this with the choose function ?
What’s the explanation given?
Initially, you have 10! ways of arranging 10 unique words. After you impose the restriction on the first three letters, you have 7 \cdot 6 \cdot 5 \cdot 7! ways of arranging 10 unique words.
Thus, \frac{7 \cdot 6 \cdot 5 \cdot 7!}{10 \cdot 9 \cdot 8 \cdot 7!} = \frac{7}{24}
Could you explain why it’s wrong to do (10C3 X 7!)/10! aside from the obvious that the answer is more than 1 ? Because I thought that since the first 3 numbers are fixed and can be any 3 numbers we would need to multiply by 10C3.
First three words in the second randomization can be anything besides the first three words in the first randomization. 10C3 doesn’t impose this restriction.
For example, if the first three words in the first randomization were {a,b,c} then these can’t be in the first three words of the second randomization. 10C3 gives you all 3 element subsets, which also includes elements with {a, something, something}, {a,b, something}, and {a,b,c}.
Moreover, if you do decide to now switch to 7C3 now, you still have to account for all permutations of a given subset. For instance, if we’re looking at {e,f,g} for the first three words in the second randomization then all permutations of this works equally well too.
So that’d leave you with 7C3 \cdot 3! = 7P3