At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Jayden and Alexa made mistakes in their calculations. Let's walk through their mistakes and provide the corrected solution step-by-step.
### Jayden's Approach
1. Jayden incorrectly simplified the combinations and treated the factorial calculations incorrectly. His attempt was:
[tex]\[\frac{6!}{2!(6-2)!} + \frac{6!}{3!(6-3)!} \neq \frac{7!}{3!(7-3)!}\][/tex]
Simplifying,
[tex]\[\frac{6!}{2! \cdot 4!} + \frac{6!}{3! \cdot 3!} \neq \frac{7!}{3! \cdot 4!}\][/tex]
2. Correct calculation of [tex]\({ }_6C_2\)[/tex] and [tex]\({ }_6C_3\)[/tex]:
[tex]\[ { }_6C_2 = \frac{6!}{2! \cdot 4!} = \frac{6 \cdot 5}{2 \cdot 1} = 15 \][/tex]
[tex]\[ { }_6C_3 = \frac{6!}{3! \cdot 3!} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20 \][/tex]
3. Sum of the calculated combinations:
[tex]\[ { }_6C_2 + { }_6C_3 = 15 + 20 = 35 \][/tex]
4. Correct calculation of [tex]\({ }_7C_3\)[/tex]:
[tex]\[ { }_7C_3 = \frac{7!}{3! \cdot 4!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 \][/tex]
Therefore, [tex]\({ }_6C_2 + { }_6C_3 = { }_7C_3\)[/tex] is indeed true as both sides simplify to 35.
### Alexa's Approach
1. Alexa's initial step was correct but her use of the least common denominator was convoluted. Here's the correction using combinations:
[tex]\[ { }_6C_2 = \frac{6!}{2! \cdot 4!} \][/tex]
[tex]\[ { }_6C_3 = \frac{6!}{3! \cdot 3!} \][/tex]
[tex]\(\Rightarrow\)[/tex]
[tex]\[ { }_6C_2 + { }_6C_3 = 15 + 20 \][/tex]
2. As before, the combined result is:
[tex]\[ 15 + 20 = 35 \][/tex]
3. Alexa’s final check:
[tex]\[ { }_7C_3 = \frac{7!}{3! \cdot 4!} \][/tex]
[tex]\(\Rightarrow\)[/tex]
[tex]\[ { }_7C_3 = 35 \][/tex]
So, once corrected, Alexa's result too shows [tex]\({ }_6C_2 + { }_6C_3 = { }_7C_3.\)[/tex]
### Conclusion
Both Jayden and Alexa made mistakes initially but when corrected, both paths show the correct result. The identity [tex]\({ }_6C_2 + { }_6C_3 = { }_7C_3\)[/tex] holds true since:
[tex]\[ 15 + 20 = 35,\][/tex]
and hence,
[tex]\[35 = 35.\][/tex]
Thus, Jayden and Alexa both arrive at the correct conclusion when their errors are corrected.
### Jayden's Approach
1. Jayden incorrectly simplified the combinations and treated the factorial calculations incorrectly. His attempt was:
[tex]\[\frac{6!}{2!(6-2)!} + \frac{6!}{3!(6-3)!} \neq \frac{7!}{3!(7-3)!}\][/tex]
Simplifying,
[tex]\[\frac{6!}{2! \cdot 4!} + \frac{6!}{3! \cdot 3!} \neq \frac{7!}{3! \cdot 4!}\][/tex]
2. Correct calculation of [tex]\({ }_6C_2\)[/tex] and [tex]\({ }_6C_3\)[/tex]:
[tex]\[ { }_6C_2 = \frac{6!}{2! \cdot 4!} = \frac{6 \cdot 5}{2 \cdot 1} = 15 \][/tex]
[tex]\[ { }_6C_3 = \frac{6!}{3! \cdot 3!} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20 \][/tex]
3. Sum of the calculated combinations:
[tex]\[ { }_6C_2 + { }_6C_3 = 15 + 20 = 35 \][/tex]
4. Correct calculation of [tex]\({ }_7C_3\)[/tex]:
[tex]\[ { }_7C_3 = \frac{7!}{3! \cdot 4!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 \][/tex]
Therefore, [tex]\({ }_6C_2 + { }_6C_3 = { }_7C_3\)[/tex] is indeed true as both sides simplify to 35.
### Alexa's Approach
1. Alexa's initial step was correct but her use of the least common denominator was convoluted. Here's the correction using combinations:
[tex]\[ { }_6C_2 = \frac{6!}{2! \cdot 4!} \][/tex]
[tex]\[ { }_6C_3 = \frac{6!}{3! \cdot 3!} \][/tex]
[tex]\(\Rightarrow\)[/tex]
[tex]\[ { }_6C_2 + { }_6C_3 = 15 + 20 \][/tex]
2. As before, the combined result is:
[tex]\[ 15 + 20 = 35 \][/tex]
3. Alexa’s final check:
[tex]\[ { }_7C_3 = \frac{7!}{3! \cdot 4!} \][/tex]
[tex]\(\Rightarrow\)[/tex]
[tex]\[ { }_7C_3 = 35 \][/tex]
So, once corrected, Alexa's result too shows [tex]\({ }_6C_2 + { }_6C_3 = { }_7C_3.\)[/tex]
### Conclusion
Both Jayden and Alexa made mistakes initially but when corrected, both paths show the correct result. The identity [tex]\({ }_6C_2 + { }_6C_3 = { }_7C_3\)[/tex] holds true since:
[tex]\[ 15 + 20 = 35,\][/tex]
and hence,
[tex]\[35 = 35.\][/tex]
Thus, Jayden and Alexa both arrive at the correct conclusion when their errors are corrected.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.