20-students-should-stand-in-5-different-rows-each-row-should-have-at-least-2-students-in-how-many-ways-can-you-arrange-them-

Question Number 124878 by mr W last updated on 06/Dec/20

20 s t u d e n t s s h o u l d s t a n d i n 5

d i f f e r e n t r o w s . e a c h r o w s h o u l d h a v e

a t l e a s t 2 s t u d e n t s . i n h o w m a n y w a y s

c a n y o u a r r a n g e t h e m ?

Answered by liberty last updated on 07/Dec/20

(1) [\underset{2}{\overset{R_{1}}{-}}, \underset{2}{\overset{R_{2}}{-}}, \underset{2}{\overset{R_{3}}{-}}, \underset{2}{\overset{R_{4}}{-}}, \underset{12}{\overset{R_{5}}{-}}] \Rightarrow \frac{5!}{4!} \times P_{2}^{20} \times P_{2}^{18} \times P_{2}^{16} \times P_{2}^{14} \times 12!

(2) [\frac{R_{1}}{2}, \frac{R_{2}}{2}, \frac{R_{3}}{2}, \frac{R_{4}}{3}, \frac{R_{5}}{11}] \Rightarrow \frac{5!}{3!} \times P_{2}^{20} \times P_{2}^{18} \times P_{2}^{16} \times P_{3}^{14} \times 11!

(3) [\frac{R_{1}}{2}, \frac{R_{2}}{2}, \frac{R_{3}}{2}, \frac{R_{4}}{4}, \frac{R_{5}}{10}] \Rightarrow \frac{5!}{3!} \times_{20} P_{2} \times_{18} P_{2} \times_{16} P_{2} \times_{14} P_{4} \times 10!

n e x t \dots

Commented by mr W last updated on 07/Dec/20

t h a n k s f o r t r y i n g!

s i n c e w e h a v e 1001 c a s e s, i {d o n}^{'} t t h i n k

w e c a n s o l v e i n t h i s w a y . b u t i k n o w

n o o t h e r w a y s h o w t o s o l v e .

Answered by mindispower last updated on 07/Dec/20

s o n c e r h e r i s a t l e a s t 2 i n r o w

w e c a n u s e p o w e r s e r i e

\frac{x^{2}}{1 - x} s t a r t e w i t h e x^{2}

{(\frac{x^{2}}{1 - x})}^{5}, t h a n f i n d c o e f i c i e n t o f x^{20}

\frac{x^{10}}{{(1 - x)}^{5}}, \frac{1}{1 - x} = Σ x^{k}

{(\frac{1}{1 - x})}^{5} = \frac{1}{4!} . \frac{\partial^{4}}{\partial x^{4}} (\frac{1}{1 - x}) = \frac{1}{4!} . \sum_{k ⩾ 0} (k + 1) (k + 2) (k + 3) (k + 4) x^{k}

\frac{x^{10}}{{(1 - x)}^{5}} = \frac{1}{4!} \sum_{k ⩾ 0} (k + 1) (k + 2) (k + 3) (k + 4) x^{k + 10}

x^{20} h a s e c o e f i c i e n t \frac{1}{4!} . (14) (13) (12) (11)

= \frac{(14)!}{4! . 10!} = C_{10}^{4}

Commented by mr W last updated on 07/Dec/20

y o u h a d a t y p o . y o u m e a n t C_{14}^{4} = 1001 .

t h e d i f f i c u l t y i s f o r e x a m p l e, i f i n t h e

f i r s t r o w t h e r e a r e 5 s t u d e n t s . t o

s e l e c t 5 f r o m 20 s t u d e n t s t h e r e a r e

C_{5}^{20} w a y s a n d t o a r r a n g e t h e s e 5

s t u d e n t s i n t h e r o w t h e r e a r e 5! w a y s .

Answered by mr W last updated on 07/Dec/20

s a y t h e n u m b e r s o f s t u d e n t s i n t h e

f i v e r o w s a r e a, b, c, d a n d e .

a + b + c + d + e = 20

2 ⩽ a, b, c, d, e ⩽ 12

t o s e l e c t a s t u d e n t s f o r r o w 1 t h e r e

a r e C_{a}^{20} w a y s, a n d t o a r r a n g e t h e s e

a s t u d e n t s i n r o w 1 t h e r e a r e a! w a y s .

t h e g e n e r a t i n g f u n c t i o n f o r r o w 1 i s

\sum_{a ⩾ 2} C_{a}^{20} a! x^{a} = \underset{a = 2}{\sum^{12}} P_{a}^{20} x^{a}

e t c .

G F = (\underset{a = 2}{\sum^{12}} P_{a}^{20} x^{a}) (\underset{b = 2}{\sum^{12}} P_{b}^{20 - a} x^{b}) (\underset{c = 2}{\sum^{12}} P_{c}^{20 - a - b} x^{c}) (\underset{d = 2}{\sum^{12}} P_{d}^{20 - a - b - c} x^{d}) (\underset{e = 2}{\sum^{12}} P_{e}^{e} x^{e})

n o w i t i s t o f i n d t h e c o e f f i c i e n t o f

t h e x^{20} t e r m i n G F w h i c h i s

\sum_{\underset{a, b, c, d, e ⩾ 2}{a + b + c + d + e = 20}} (P_{a}^{20} P_{b}^{20 - a} P_{c}^{20 - a - b} P_{d}^{20 - a - b - c} P_{e}^{e})

Commented by mindispower last updated on 08/Dec/20

i s e e w h a t y o u m e a n b y y o u r q u a t i o n

s o r r y i m n o t i n e n g l i s h i s t h e p l a c e i n r o w b y s o m o e n e d e p e n d ?

Commented by mr W last updated on 08/Dec/20

t h e f i v e r o w s a r e d i s t i n c t .

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