Question

Medal. Please Help! Jayden and Alexa were asked if 6^C^2 + 6^C^3 = 7^C^3 is true. Their calculations showed it was not true, but they both made mistakes in their calculations. Explain and correct their mistakes.

Jayden used`` the formula and simplified.
https://media.glynlyon.com/g_alg02_ccss_2013/5/jayden_calculation.gif

Alexa added the fractions by using the LCD (least common denominator).
https://media.glynlyon.com/g_alg02_ccss_2013/5/alexa_calculation.gif

Answer 1

instead of addition use multiplication,e.g,2!*4!=2*1*4*3*2*1

Answer 2

I am new to this, I really don't understand that much

Answer 3

\[6C2+6C3=\frac{ 6*5 }{ 2*1 }+\frac{ 6*5*4 }{ 3*2*1}=\frac{ 6*5 }{ 2*1 }\left( 1+\frac{ 4 }{ 3 } \right)\]
\[=\frac{ 6*5 }{2*1 }*\frac{ 7 }{ 3 }=\frac{ 7*6*5 }{ 3*2*1 }=7C3\]

Answer 4

How would I explain it?

Answer 5

in their way
\[6C2=\frac{ 6*5 }{ 2! }=\frac{ 6*5 }{ 2! }*\frac{ 4! }{ 4! }=\frac{ 6! }{ 2!*4!}\]

Answer 6

\[6c2+6c3=\frac{ 6! }{ 2!*4! }+\frac{ 6! }{ 3!*3! }\]
\[=\frac{ 6! }{ 2!*4*3!}+\frac{ 6! }{ 3*2!*3! }\]
\[=\frac{ 6! }{ 3!*2!}\left( \frac{ 1 }{ 4} +\frac{ 1 }{ 3 }\right)=\frac{ 6! }{ 3!*2! }\left( \frac{ 3+4 }{ 12 } \right)\]
\[=\frac{ 6*5*4*3*2*1*7 }{3*2*1*2*1*12 }=\frac{ 7*6*5 }{ 3*2*1 }=7C3\]

Answer 7

Thank you. Should I write what you just wrote for my answer or everything and that what you just wrote?

Answer 8

they are two ways of calculating the whole problem.
Jayden added but there should be multiplication.
Alexa wrote 2!4!=4*3!3! which is wrong
It should be 2*4*3!

Answer 9

correction it should be 2!*4*3! or 2*4*3!
2!=2*1=2

Answer 10

Ok thanks. can you help me with the second part? Use the Alexa’s method. Add the first two terms using the LCD (least common denominator). Write the third term as a fraction. Confirm that the relationship is true.

Answer 11

second method is what you want.

Answer 12

second method?

Answer 13

\[6c2+6C3=\frac{ 6! }{ 2!4! }+\frac{ 6! }{ 3!3! }=\frac{ 6*5*4*3*2*1 }{ 2*1*4*3*2*1 }+\frac{ 6*5*4*3*2*1 }{ 3*2*1*3*2*1 }\]
\[=\frac{ 6*5*4*3*2*1*3+6*5*4*3*2*1*4 }{ 4*3*2*1*3*2*1 }\]
\[=\frac{ 7*6*5*4*3*2*1 }{ 3*2*1*4*3*2*1}=\frac{ 7! }{ 3!4! }=7C3\]

Answer 14

Thank you for the help.

Answer 15

yw