anonymous
  • anonymous
how many different string can be made from the word PEPPERCORN when all letters are used and such strings do not contain the substring CON?
Meta-math
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
you can rearrange PEPPERCORN in 10! ways. then, we can consider CON as one letter, lets say Z , so instead of PEPPERCORN, we'll have PEPPERRZ, we can rearrange this in 8! ways. so I think the answer is 10!-8!
Kainui
  • Kainui
Almost right, but not quite. See, peppercorn has 10 letters in it, so it is 10!, however you have to divide out all the ways in which you can rearrange the letters that are being reused, such as P, E, and R. So P appears 3 times, E twice, and R twice so \[\frac{ 10! }{ 3!*2!*2! }\] That tells us how many different arrangements you can have, but now we need to get rid of all the occurrences of "CON". Simply enough, we can just look at the ten letter word and see by counting (or more complicated factorial work) that you can put CON in only 8 possible places, so just subtract 8 from the factorial answer above: \[\frac{ 10! }{ 3!*2!*2! }-8\] I believe that's the right answer, but I'm human too!
Kainui
  • Kainui
Wait, subtracting 8 was ridiculous lol. It too is a factorial that must be calculated similarly I don't know what I was thinking.

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RadEn
  • RadEn
the possibles for word "PEPPERRZ" is 8!/(3!*2!*2!), @Kainui
Kainui
  • Kainui
Wait, that's not the final answer though.
Kainui
  • Kainui
So that tells you all the occurrences that will contain CON in it, while the other calculation I did will tell how many total words you can make. Seems we can just subtract the two and get the final answer: \[\frac{ 10! }{ 3!*2!*2! }-\frac{ 8! }{ 3!*2!*2! }\]
RadEn
  • RadEn
yeah, i meant the finally it will be : 10!/(3!*2!*2!) - 8!/(3!*2!*2!)

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