At a charity benefit with 22 people in attendance, three $50 gift cert

1.	At a charity benefit with 22 people in attendance, three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, in how many different ways can the gift certificates be awarded?
Answer» $\huge{\underline{\mathtt{\red {A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$ In the Isosceles TRIANGLE ABC, ∠BAC = 30° So, ∠CBA also = 30° = X (Opposite side in a Isosceles ∆) _____________________________________ If ∠BAC & ∠CBA = 30° Then, ∠BAC + ∠CBA + ∠ACB = 180° (Angle Sum Prop. of a Triangle) ➠ 30° + 30° + ∠ACB = 180° ➠ ∠ACB = 180° - 60° ➠ ∠ACB = 120° If ACB = 120° Then, ∠BCD = 180° - 120° ∠BCD = z = 60° ___________________________________ If z = 60° Then, Y also = 60° (OPP. Sides in an Isosceles ∆ ) ____________________________________ OVERALL, x = 30° y = 60° z = 60° . $⠀⠀⠀\orange{─══━━━━✥◈✥━━━━══─}$ $⠀⠀⠀⠀⠀⠀\bold\red{ℜびのモ \:みモℜロ}$ $⠀⠀⠀\purple{─══━━━━✥◈✥━━━━══─}$ ⠀

At a charity benefit with 22 people in attendance, three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, in how many different ways can the gift certificates be awarded?

Answer»

$\huge{\underline{\mathtt{\red {A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

In the Isosceles TRIANGLE ABC,

∠BAC = 30°

So,

∠CBA also = 30° = X (Opposite side in a Isosceles ∆)

_____________________________________

If ∠BAC & ∠CBA = 30°

Then,

∠BAC + ∠CBA + ∠ACB = 180° (Angle Sum Prop. of a Triangle)

➠ 30° + 30° + ∠ACB = 180°

➠ ∠ACB = 180° - 60°

➠ ∠ACB = 120°

If ACB = 120° Then,

∠BCD = 180° - 120°

∠BCD = z = 60°

___________________________________

If z = 60°

Then,

Y also = 60° (OPP. Sides in an Isosceles ∆ )

____________________________________

x = 30°

y = 60°

z = 60°

$⠀⠀⠀\orange{─══━━━━✥◈✥━━━━══─}$

$⠀⠀⠀⠀⠀⠀\bold\red{ℜびのモ \:みモℜロ}$

$⠀⠀⠀\purple{─══━━━━✥◈✥━━━━══─}$ ⠀