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.