𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Prove 20!/6!14! + 20!/7!13! = 21!/7!14!

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

We need to show that the left-hand side (LHS) is equal to the right-hand side (RHS).

LHS = 20! / (6! 14!) + 20! / (7! 13!)

To add these fractions, we need a common denominator. The least common multiple of 6! 14! and 7! 13! is 7! 14! (since 7! = 7 × 6! and 14! = 14 × 13!).

Rewrite the first term with the common denominator:

20! / (6! 14!) = (20! × 7) / (6! × 7 × 14!) = (7 × 20!) / (7! 14!)

Rewrite the second term with the common denominator:

20! / (7! 13!) = (20! × 14) / (7! × 13! × 14) = (14 × 20!) / (7! 14!)

Now add the two terms:

LHS = (7 × 20!) / (7! 14!) + (14 × 20!) / (7! 14!)

LHS = (7 × 20! + 14 × 20!) / (7! 14!)

Factor out 20! from the numerator:

LHS = (20! × (7 + 14)) / (7! 14!)

LHS = (20! × 21) / (7! 14!)

Recall that 21! = 21 × 20!

Substitute this into the numerator:

LHS = 21! / (7! 14!)

This is equal to the RHS.

Therefore, 20! / (6! 14!) + 20! / (7! 13!) = 21! / (7! 14!) is proven.