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Question Number 119390 by mr W last updated on 24/Oct/20

w e h a v e 15 d i f f e r e n t m a t h e m a t i c s

b o o k s, 10 d i f f e r e n t p h y s i c s b o o k s a n d

12 d i f f e r e n t c h e m i s t r y b o o k s . w e s h o u l d

c h o o s e 6 b o o k s s u c h t h a t t h e y c o n t a i n

a l l t h r e e k i n d s o f b o o k s .

i n h o w m a n y w a y s c a n w e d o t h i s ?

Answered by bemath last updated on 24/Oct/20

(1) (4, 1, 1) \Rightarrow \frac{P_{4}^{15} \times P_{1}^{10} \times P_{1}^{12}}{3!}

(2) (3, 2, 1) \Rightarrow \frac{P_{3}^{15} \times P_{2}^{10} \times P_{1}^{12}}{3!}

(3) (3, 1, 2) \Rightarrow \frac{P_{3}^{15} \times P_{1}^{10} \times P_{2}^{12}}{3!}

(4) (2, 3, 1) \Rightarrow \frac{P_{2}^{15} \times P_{3}^{10} \times P_{1}^{12}}{3!}

(5) (2, 2, 2) \Rightarrow \frac{P_{2}^{15} \times P_{2}^{10} \times P_{2}^{12}}{3!}

(6) (2, 1, 3) \Rightarrow \frac{P_{2}^{15} \times P_{1}^{10} \times P_{3}^{12}}{3!}

(7) (1, 4, 1) \Rightarrow \frac{P_{1}^{15} \times P_{4}^{10} \times P_{1}^{12}}{3!}

(8) (1, 3, 2) \Rightarrow \frac{P_{1}^{15} \times P_{3}^{10} \times P_{2}^{12}}{3!}

(9) (1, 2, 3) \Rightarrow \frac{P_{1}^{15} \times P_{2}^{10} \times P_{3}^{12}}{3!}

(10) (1, 1, 4) \Rightarrow \frac{P_{1}^{15} \times P_{1}^{10} \times P_{4}^{12}}{3!}

T o t a l l y = (1) + (2) + (3) + \dots + (10)

Commented by mr W last updated on 24/Oct/20

w h y d o y o u u s e P_{k}^{n}, n o t C_{k}^{n} ?

w h e n w e c h o o s e 3 m a t h e m a t i c s b o o k s,

w e {d o n}^{'} t a r r a n g e t h e m .

Commented by bemath last updated on 24/Oct/20

b e c a u s e t h e b o o k s a r e d i f f e r e n t

Commented by mr W last updated on 24/Oct/20

t o c h o o s e 3 f r o m 15 d i f f . m a t h e m a t i c s

b o o k s t h e r e a r e C_{3}^{15} w a y s, n o t P_{3}^{15}

Commented by bemath last updated on 24/Oct/20

s i r t h e a r r a n g e m e n t A B C n o t s a m e t o B A C

Commented by mr W last updated on 24/Oct/20

w e s e l e c t 3 m a t h e m a t i c s b o o k s A, B, C,

{d o n}^{'} t a r r a n g e t h e m . s o A B C, A C B,

B A C e t c . a r e t h e s a m e .

Answered by ebi last updated on 24/Oct/20

l e t A = n u m b e r o f w a y s c h o o s i n g 6 b o o k s

f r o m 3 k i n d s o f b o o k s A N D c o n t a i n

a l l 3 k i n d s o f b o o k s .

(m a t h s, p h y, c h e m)

(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1),

(2, 1, 3), (2, 2, 2), (2, 3, 1),

(3, 1, 2), (3, 2, 1),

(4, 1, 1)

A

=^{15} C_{1}^{10} C_{1}^{12} C_{4} +^{15} C_{1}^{10} C_{2}^{12} C_{3} +

C_{1}^{10} C_{3}^{12} C_{2} +^{15} C_{1}^{10} C_{4}^{12} C_{1} +

C_{2}^{10} C_{1}^{12} C_{3} +^{15} C_{2}^{10} C_{2}^{12} C_{2} +

C_{2}^{10} C_{3}^{12} C_{1} +^{15} C_{3}^{10} C_{1}^{12} C_{2} +

C_{3}^{10} C_{2}^{12} C_{1} +^{15} C_{4}^{10} C_{1}^{12} C_{1}

= (15 \times 10 \times 495) + (15 \times 45 \times 220) +

(15 \times 120 \times 66) + (15 \times 210 \times 12) +

(105 \times 10 \times 220) + (105 \times 45 \times 66) +

(105 \times 120 \times 12) + (455 \times 10 \times 66) +

(455 \times 45 \times 12) + (1365 \times 10 \times 12)

= 74250 + 148500 + 118800 + 37800 +

231000 + 311850 + 151200 + 300300 +

245700 + 163800

= 1783200

Commented by mr W last updated on 24/Oct/20

c o r r e c t a n s w e r, t h a n k s!

Commented by mr W last updated on 24/Oct/20

t h i s i s a l s o t h e m e t h o d i t h o u g h t o f .

b u t i f w e h a v e 4 o r m o r e k i n d s o f

b o o k s, t h i s m e t h o d i s v e r y t o u g h .

d o y o u h a v e a n y o t h e r i d e a s ?

Commented by mr W last updated on 25/Oct/20

p l e a s e t r y Q 119515

Commented by ebi last updated on 25/Oct/20

i t h i n k s o l v i n g p r o b a b i l i t y p r o b l e m s

a r e q u i t e h a r d, a s h a v e t o l i s t a l l t h e

p o s s i b l e o u t c o m e s a n d c a l c u l a t e t h e m

o n e b y o n e . i s t i l l t h i n k i n g t o

s o l v e i t i n e a s i e r w a y . b u t, t h i s m e t h o d

i t h i n k i s t h e b e s t w a y .

Commented by mr W last updated on 25/Oct/20

t h a n k s!

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