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the number of words in which the letters of the word DECISIONS be arranged so that letter N is somewhere in between I’s is k! / 12 where k is______________ int type qs..

the number of words in which the letters of the word DECISIONS be arranged so that letter N is somewhere in between I’s is k! / 12 where k is______________
int type qs..

Grade:12

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find the answer to your question below.
Sorry for the previous misunderstanding as i have misinterpreted the question.

Word – DECISIONS
S & I in the word repeat two times.
Since we need to fix N b/w two I’s.
We have something like this
(DECSOS)(INI).........(1)
This is arrangement to fix N b/w two I’s.
Now we treat (INI) as one unit (or one letter).
Number of ways to arrange the in (1)
There are seven letter in which S is repeating two times.
\frac{7!}{2!}
Number of ways of arranging INI with N b/w two I’s
\frac{2!}{2!}
Total number of ways
\frac{7!}{2!}.\frac{2!}{2!} = \frac{7!}{2!}
\frac{7!}{2!} = \frac{9.8.7!}{9.8.2!} = \frac{9!}{144} = \frac{9!}{(12)^{2}}
Now, (INI) can be arrange in six similar type of arrangement.
(INI)DECSOS,D(INI)ECSOS,DE(INI)CSOS,DEC(INI)SOS,DECS(INI)OS,DECSO(INI)S
Total no. of ways
\frac{7.9!}{(12)^{2}}….........(2)

2nd Part of the problem:
We have seven spaces free around (DECSOS) to arrange INI
.D.E.C.S.O.S.
. are free spaces.
We have to choose three places from them to arrange INI.
No. of ways
7C3.(2!/2!) = 7C3
No. of ways to arrange DECSOS
\frac{6!}{2!}
Total no. of ways
7C3(6!/2!)
\frac{5.9!}{(12)^{2}}…................(3)
Total ways
(2) + (3)
\frac{7.9!}{(12)^{2}}+\frac{5.9!}{(12)^{2}}
\frac{12.9!}{(12)^{2}} = \frac{9!}{12}
On comparing, K!/12 with 9!/12, we get
k = 9
SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find the answer to your question below
There are 9 letters so there are 7 cases
case 1: i _ _ _ _ _ _ _ i
here no of ways =7!/2!
i _ _ _ _ _ _ i_
no of ways=6*6!/2!
i _ _ _ _ _ i _ _
no of ways=6*5*5!/2!
similarly we will different ways get as i moves forward every time by one position as
6*5*4*4!/2! , 6*5*4*3*3!/2! ,6*5*4*3*2*2!/2! ,6*5*4*3*2*1!/2! till the case ini_ _ _ _ _ _ occurs
case 2:_i _ _ _ _ _ _ i
by writing ways similarly as above we get
6*6!/2!,6*5*5!/2!,6*5*4*4!/2! , 6*5*4*3*3!/2! ,6*5*4*3*2*2!/2! ,6*5*4*3*2*1!/2! till the case _ini _ _ _ _ _ occurs
case3:_ _i_ _ _ _ _ i
by writing ways similarly as above we get
6*5*5!/2!,6*5*4*4!/2! , 6*5*4*3*3!/2! ,6*5*4*3*2*2!/2! ,6*5*4*3*2*1!/2! till the case_ _ini _ _ _ _ occurs
case 4:_ _ _i _ _ _ _ i
by writing ways similarly as above we get
6*5*4*4!/2! , 6*5*4*3*3!/2! ,6*5*4*3*2*2!/2! ,6*5*4*3*2*1!/2! till the case_ _ _ini_ _ _ occurs
case 5:_ _ _ _i _ _ _ i
by writing ways similarly as above we get
6*5*4*3*3!/2! ,6*5*4*3*2*2!/2! ,6*5*4*3*2*1!/2! till the case_ _ _ _ ini _ _ occurs
case 6:_ _ _ _ _i _ _ i
by writing ways similarly as above we get
6*5*4*3*2*2!/2! ,6*5*4*3*2*1!/2! till the case_ _ _ _ _ini_ occurs
case 7:_ _ _ _ _ _i ni
by writing ways similarly as above we get
6*5*4*3*2*1!/2! till the case_ _ _ _ _ _ini occurs
By adding all the above cases we get
(7*6!/2!)+(2*6*6!/2!)+(5*3*6!/2!)+(4*4*6!/2!)+(3*5*6!/2!)+(2*6*6!/2!)+(7*6!/2!)=84*6!/2=9!/12
hence the value of k=9

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