How many arrangements can be made by the letters of word DEFINITION

Question

How many arrangements can be made by the letters of word DEFINITION if the letters I do not occupy the first or last place?

A. 20,160
B. 40,320
C. 70,560
D. 141,120
E. 282,240

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chetan2u · Accepted Answer

These questions require us to first analyse the letters. 
I:3
N:2
D,E,F,T,O: 1 each

Total ways without any restrictions  =  \frac{10!}{3!2!}=10*9*8*7*6*5*2=302400

Now, let us check for the ways with restrictions 
 I at first spot : The remaining 9 can be put in  \frac{9!}{2!2!}=90720 
Similarly  for I at last spot : 90720.

However the ways when I is at both first and the last spot is being subtracted twice, once in each case. We will, therefore, have to add these back once. 
 I at the first and the last spot:  Remaining 8 can be placed in  \frac{8!}{2!}=20160 
So 2*90720-20160=161280

Total ways with restrictions  = 302400-161280=141120

D

rvgmat12

CrackverbalGMAT · Answer

Number of words where I is not in the 1st or last = Total words - Words where I occupies the 1st and last place

DEFINITION has 10 letters with 3 I's, 2 N's and 1 each of D, E, F, T and O.

Total words = 10! / 3!2!

Words where I comes in the 1st and the last place.

2 I's can be placed in these 2 positions in 1 way. The remaining 8 letters can be arranged in 8!/2! ways (these contain 2 N's)

Therefore Total number of words =  \frac{10!}{3!2!} - \frac{8!}{2!} = \frac{10 * 9 * 8!}{6 * 2} - \frac{8!}{2} = \frac{8!}{2}(15 - 1)

= 40320 * 7 = 282240

Option E

Arun Kumar

ShaikhMoice · Answer

Bunuel Please help with the solution.
Thanks in Advance

GmatKnightTutor · Answer

D) (7 times 6 times 8!)/3! 2!

rvgmat12 · Answer

Dear  chetan2u , could you please solve this ? Thanks

jrk23 · Answer

Total ways (10!/3!*2!)- When I come on First Place(9!/2!*2!)-when i comes on last position(9!/2!*2!) + when I comes on both place(8!/2!)= 141120

chetan2u · Answer

You will have to subtract ways wherein I is at first place but not last => 
1) Last position any of 5 single letters, and then remaining can be filled in 8!/2!2! ===>5*8!/2!2!=5*10080=50400
2) Last position by N, and then remaining can be filled in 8!/2! ===>1*8!/2!=1*20160
Total = 50400+20160=70560

Similarly we will have 70560 ways when I is at last place but not first

Our answer =282240-2*70560=282240-141120=141120

Lipun · Answer

Another approach to solve this -

The middle 8 places are the only options for arranging 3 Is.

The other 5 characters to fill up these 8 places can be chosen in 7C5 ways.

Total ways of arranging these 8 characters = 7C5 * (8!/3!) = 7C5 * 8P3

The remaining two characters from the list of 7 will occupy the first and last places. These can be further arranged in 2! ways.

Also, 'N' has a cardinality of 2. Thus, we need to divide by 2! to mitigate the repetitions of 'N'.

Total number of possible arrangements = (7C5 * 8P3 * 2!) / 2! = 7C5 * 8P3 = 141120

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How many arrangements can be made by the letters of word DEFINITION

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