Question & Answer
QUESTION

8-digit numbers are formed using the digits 1,1,2,2,2,3,4,4. The number of such numbers in which the odd digits do not occupy odd places, is:
(A) 160
(B) 120
(C) 60
(D) 48

ANSWER Verified Verified

Hint: Whenever you come up with this type of problem then first divide it into different cases. And then find ways for different cases then the total number of ways will be the sum of ways of all cases.


Complete step by step answer:


As, we all know that,

Number of arrangements of like and alike objects is such that,

The number of arrangements that can be formed using n objects out of which r,q and p are identical objects.

Then, total number of arrangements will be $\dfrac{{n!}}{{p!q!r!}}$

 

So here as given in the question we have to arrange the given numbers in such a way that the all the odd places of the 8 - digit number will not be occupied by odd numbers.

So, from the given numbers there are three odd numbers and that are 1,1,3 


And there are five even numbers and that are 2,2,2,4,4 


And as we know that there are 4 odd places in a 8 - digit number

We have to fill that odd places with even numbers and then fill the left even places with other numbers

So, for that we have to choose four numbers out of five given even numbers.


As, there are only four odd places

And then fill all even places with three odd numbers and one even number that is left.

So, for doing this there can be two case


Case 1: odd places filled by (2,2,4,4) and even places filled by (1,1,3,2)


So, as from the above mentioned arrangements of like and alike things


Number of ways $= \left( {\left( {\dfrac{{4!}}{{2!*2!}}} \right){\text{for odd places}}} \right){\text{* }}\left( {\left( {\dfrac{{4!}}{{2!}}} \right){\text{for even places}}} \right)$


So, number of ways for case 1 $= \dfrac{{24}}{4}*\dfrac{{24}}{2} = 72{\text{ ways}}$


Case 2: odd places filled by (2,2,2,4) and even places filled by (1,1,3,4)

So, as from the above mentioned arrangements of like and alike things

Number of ways $= \left( {\left( {\dfrac{{4!}}{{3!}}} \right){\text{for odd places}}} \right){\text{* }}\left( {\left( {\dfrac{{4!}}{{2!}}} \right){\text{for even places}}} \right)$


So, number of ways for case 2 $= \dfrac{{24}}{6}*\dfrac{{24}}{2} = 48{\text{ ways}}$


To total number of ways will be 72 + 48 = 120 ways 


Hence correct option is B


NOTE: Identical objects can be arranged in 1 way, and to find the arrangement of n different things of which some are identical then we divide by the arrangement of identical things assuming the identical things to be different.