How many different signals can be transmitted by arranging \[2\] red, \[3\] yellow and \[2\] green flags on a pole? (Assume that all the \[7\] flags are used to transmit a signal).
Answer
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Hint:In order to determine the different ways signals can be transmitted.The signals can be transmitted by arranging \[2\]red, \[3\] yellow and \[2\] green flags on a pole. We use the factorial method to solve the problem. The product of a whole number \['n'\] with every whole number until \[1\] is called the factorial. Finally, we obtain the required solution.
Complete step by step answer:
In this given problem, the statement includes the given as \[2\] red, \[3\] yellow and \[2\] green flags to be in a pole. Here, We use factorial method for solving the given problem,
The total number of ways to arrange if all are different = \[7!\]
But here \[3\] yellows are the same and \[2\] reds and \[2\] green are also the same.
Thus, the number of ways is \[ = \dfrac{{7!}}{{3! \times 2! \times 2!}}\]
Expanding the factorial terms to find the value of it.
\[\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{(3 \times 2 \times 1) \times (2 \times 1) \times (2 \times 1)}}\]
By simplify the multiplication of the fraction, then
\[\dfrac{{5040}}{{(6) \times (2) \times (2)}}\]
Multiplying on the values to find the value of the number of ways,
\[\dfrac{{5040}}{{24}}\]
The number of ways is \[ = 210\]
Thus, the number of ways to arrange the flags in the pole is \[210\]. The solution for the above question in which they asked for different signals to be transmitted by arranging \[2\] red, \[3\] yellow, and \[2\] green flags on the pole.Based on the calculation using factorial ways the answer to be found as it can be arranged in \[210\] ways. Therefore, \[210\] different signals can be transmitted by arranging \[2\] red, \[3\] yellow and \[2\] green flags on a pole.
Note: Some of the interesting facts about Factorials they are:
-The number of zeros at the end of n! is roughly \[\dfrac{n}{4}\]
-\[70!\] is the smallest factorial larger than a googol.
-The sum of the reciprocals of all factorials is \[e\].
-Factorials can be extended to fractions, negative numbers and complex numbers by the gamma function.
Complete step by step answer:
In this given problem, the statement includes the given as \[2\] red, \[3\] yellow and \[2\] green flags to be in a pole. Here, We use factorial method for solving the given problem,
The total number of ways to arrange if all are different = \[7!\]
But here \[3\] yellows are the same and \[2\] reds and \[2\] green are also the same.
Thus, the number of ways is \[ = \dfrac{{7!}}{{3! \times 2! \times 2!}}\]
Expanding the factorial terms to find the value of it.
\[\dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{(3 \times 2 \times 1) \times (2 \times 1) \times (2 \times 1)}}\]
By simplify the multiplication of the fraction, then
\[\dfrac{{5040}}{{(6) \times (2) \times (2)}}\]
Multiplying on the values to find the value of the number of ways,
\[\dfrac{{5040}}{{24}}\]
The number of ways is \[ = 210\]
Thus, the number of ways to arrange the flags in the pole is \[210\]. The solution for the above question in which they asked for different signals to be transmitted by arranging \[2\] red, \[3\] yellow, and \[2\] green flags on the pole.Based on the calculation using factorial ways the answer to be found as it can be arranged in \[210\] ways. Therefore, \[210\] different signals can be transmitted by arranging \[2\] red, \[3\] yellow and \[2\] green flags on a pole.
Note: Some of the interesting facts about Factorials they are:
-The number of zeros at the end of n! is roughly \[\dfrac{n}{4}\]
-\[70!\] is the smallest factorial larger than a googol.
-The sum of the reciprocals of all factorials is \[e\].
-Factorials can be extended to fractions, negative numbers and complex numbers by the gamma function.
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