Explain how to solve ":'

Dear Student,

T o t a l n u m b e r o f w a y s t o d r a w 7 t i c k e t s i n 4 d r a w i n g s = 7^{4} = 2401 F v a o u r a b l e n u m b e r o f w a y s s u c h t h a t i n 4 d r a w i n g s t h e s u m o f n u m b e r s i s 8 = c o e f f i c i e n t o f x^{8} i n {(1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})}^{4} 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} i s a G P w i t h c o m m o n r a t i o x a n d f i r s t t e r m 1 s u m o f G P 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} = \frac{1 \times (1 - x^{7})}{(1 - x)} H e n c e w e n e e d t o f i n d c o e f f i c i e n t o f x^{8} i n {(\frac{1 - x^{7}}{1 - x})}^{4} \Rightarrow {(1 - x^{7})}^{4} \times {(1 - x)}^{- 4} \Rightarrow [1 - 4 x^{7} + x^{28} - 4 x^{21} + 6 x^{14}] \times {(1 - x)}^{- 4} \Rightarrow [1 - 4 x^{7} + . . a s h i g h e r p o w e r s t h a n 8] \times {(1 - x)}^{- 4} n o w b i n o m i a l e x p a n s i o n o f {(1 - x)}^{- n} u \sin g t a y l o r s e r i e s = 1 + n x + \frac{n \times (n + 1)}{2!} x^{2} + \frac{n \times (n + 1) (n + 3)}{3!} x . . . . \infty \Rightarrow {(1 - x)}^{- n} = \sum_{k = 0}^{\infty} {}^{k + 3}C_{3} \times x^{k} h e n c e {(1 - x)}^{- 4} w e g e t = {}^{8 + 3}C_{3} \times x^{8} + {}^{1 + 3}C_{3} \times x^{1} + a n d o t h e r p o w e r s w h i c h a r e n o t r e q u i r e d \Rightarrow [1 - 4 x^{7} + . . a s h i g h e r p o w e r s t h a n 8] \times [{}^{8 + 3}C_{3} \times x^{8} + {}^{1 + 3}C_{3} \times x^{1} + o t h e r p o w e r s] c o e f f i c i e n t o f 8 = {}^{8 + 3}C_{3} \times x^{8} - 4 \times {}^{1 + 3}C_{3} = {}^{11}C_{3} - 4 \times {}^{4}C_{3} = 165 - 16 = 149 H e n c e p r o b a b i l i t y = \frac{149}{2401}

Regards,

View Full Answer