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goblinko [34]
1 year ago
10

) a mathematical organization is producing a set of commemorative license plates. each plate contains a sequence of five charact

ers chosen from the four letters in aime and the four digits 2007. no character may appear in a sequence more times than it appears among the four letters in aime or the four digits in 2007. a set of plates in which each possible sequence appears exactly once contains n license plates. find n
Mathematics
1 answer:
forsale [732]1 year ago
8 0

The correct answer or possible sequence is the value of N/10 is 372.

The letters can be selected from the words "AIME" or "2007."

The only number that can be repeated twice is 0; there are seven different characters that can be chosen.

There are two potential scenarios.

Here is the first instance:

There are 7!/(7-5) potential sequences if 0 appears 0 or 1 times in the sequence.

There are 2520 possible sequences if 0 appears 0 or 1 times in the sequence.

Here is the second example:

If 0 occurs twice in the sequence, there must be a 0 in ⁵C₂. positions.

If 0 occurs twice in the sequence, 10 spots must be filled with 0.

The remaining three characters can be arranged in 6!/(6-3)! different ways.

There are 120 different arrangements for the final three characters.

In the second scenario, there are a total of 1200 possible possibilities, or 10*120.

There are 2520 methods in total plus 1200.

There are 3720 different ways in all.

N = 3720

We must ascertain the value of (N/10)

⇒ N/10 = 3720/10

⇒ N/10 = 372

To find more about the 'sequence' related questions

visit- brainly.com/question/28931529

#SPJ4

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Answer:

\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{r_{1} }{r_{2}}) ^{2}

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Step-by-step explanation:

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\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81\pi }{36\pi }

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