Find-the-number-of-possible-arrangements-of-the-letters-in-the-word-PENCILS-if-a-E-is-next-to-I-b-E-comes-before-I-c-there-are-three-letters-between-E-and-I-

Question Number 119397 by bemath last updated on 24/Oct/20

F i n d t h e n u m b e r o f p o s s i b l e a r r a n g e m e n t s

o f t h e l e t t e r s i n t h e w o r d P E N C I L S i f

(a)^{'} E^{'} i s n e x t t o^{'} I^{'}

(b) E c o m e s b e f o r e I

(c) t h e r e a r e t h r e e l e t t e r s b e t w e e n E a n d I

Commented by liberty last updated on 24/Oct/20

(c) E - - - I - - \Rightarrow 2! \times C_{3}^{5} \times 3! \times 2! = 240

- E - - - I - \Rightarrow 240

- - E - - - I \Rightarrow 240

T o t a l l y = 3 \times 240 = 720

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

(a)

2! \times 6! = 1440

(b)

6! = 720

(c)

X X E X X X I

X E X X X I X

E X X X I X X

\Rightarrow 3 \times 2! \times 5! = 720

Answered by liberty last updated on 24/Oct/20

(a) E I - - - - - \Rightarrow 5! = 120

- E I - - - - \Rightarrow C_{1}^{5} \times 4! = 120

- - E I - - - \Rightarrow 2! \times C_{2}^{5} \times 3! = 120

- - - E I - - \Rightarrow 3! \times C_{3}^{5} \times 2! = 120

- - - - E I - \Rightarrow 4! \times C_{4}^{5} \times 1! = 120

- - - - - E I \Rightarrow 5! = 120

T o t a l l y = 6 \times 120 = 720

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

“ E i s n e x t t o I^{″}

m e a n s “ {E I}^{″} a n d a l s o “ {I E}^{″} .

Commented by bemath last updated on 24/Oct/20

t h a n k y o u

Facebook Tweet Pin