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Jayden and Alexa were asked if [tex]${ }_6 C_2+{ }_6 C_3={ }_7 C_3$[/tex] is true. Their calculations showed it was not true, but they both made mistakes in their calculations. Explain and correct their mistakes.

Jayden's Calculation:

Step 1: [tex]\frac{6!}{2!4!} + \frac{6!}{3!3!} \neq \frac{7!}{3!4!}[/tex]

Step 2: [tex]\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 1} + \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} \neq \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 4 \cdot 3 \cdot 2 \cdot 1}[/tex]

Step 3: [tex]\frac{720}{2 \cdot 24} + \frac{720}{6 \cdot 6} \neq \frac{7 \cdot 720}{6 \cdot 24}[/tex]

Step 4: [tex]\frac{720}{48} + \frac{720}{36} \neq \frac{5040}{144}[/tex]

Step 5: [tex]15 + 20 \approx 35[/tex]

Alexa's Calculation:

Step 1: [tex]\frac{6!}{2!4!} + \frac{6!}{3!3!}[/tex]

Step 2: [tex]\frac{6!}{2! \cdot 4!} + \frac{6!}{3! \cdot 3!}[/tex]

Step 3: [tex]\frac{5 \cdot 6}{2! \cdot 4 \cdot 3!} + \frac{5 \cdot 6}{3! \cdot 3 \cdot 2!}[/tex]

Step 4: [tex]\frac{5 \cdot 6}{2 \cdot 4 \cdot 6} + \frac{5 \cdot 6}{6 \cdot 3 \cdot 2}[/tex]

Step 5: [tex]\frac{7!}{3! \cdot 4!} \neq \frac{7 \cdot 6!}{4 \cdot 3! \cdot 3!}[/tex]

Step 6: [tex]\frac{5 \cdot 6}{4 \cdot 6} \times \frac{7 \cdot 6}{4 \cdot 6}[/tex]

Correct Solution:

Using the combination formula [tex]{}_nC_r = \frac{n!}{r!(n-r)!}[/tex]:

[tex]{}_6C_2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = 15[/tex]

[tex]{}_6C_3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = 20[/tex]

[tex]{}_7C_3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = 35[/tex]

So,

[tex]{}_6C_2 + {}_6C_3 = 15 + 20 = 35 = {}_7C_3[/tex]

Therefore, the original statement is true: [tex]${ }_6 C_2+{ }_6 C_3={ }_7 C_3$[/tex]. Jayden and Alexa's errors were in their factorial simplifications and arithmetic operations.

Sagot :

GinnyAnswer
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.
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