Bob is about to hang his 8 shirts in the wardrobe. He has four different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to one another? Choose the most appropriate option:	A) 764	B) 864	C) 964	D) 1064	E) None of these
Explanation :   Total number of 8-perms of 8 shirts, recognising identical pairs, is:=8!2!4=2520In order to find the number of perms with no adjacency we subtract from this the number of perms with 4 pairs adjacent, 3 pairs adjacent, 2 pairs adjacent and 1 pair adjacent.The following four intermediate results aid explanation of the final calculation :Let the shirts be called A, A, B, B, C, C, D, D and let {A,.....} be the set of 8-perms that have only the styles named in the argument list adjacent to each other. So, for instance, {B,D} is the set of perms that contain BB and DD but not AA and CC. Let s{A,....} be the number of perms in {A,...}(i) s{A,B,C,D} = 24. This is because we can imagine each pair as one object and simply permute the four twin objects amongst themselves in 4! ways to give {A,B,C,D}.(ii) s{A,B,C} = 36. To see this we first permute the 5 objects AA, BB, CC, D, D. This can be done in5!2!=60 ways but includes {A,B,C,D}. Hence s{A,B,C} =5!2!−24(iii) s{A,B}=84. As before, we first permute the 6 objects AA, BB, C, C, D, D. This can be done in 6!/(2!)^2 ways but includes {A,B,C}, {A,B,D} and {A,B,C,D}.Hence s{A,B} =6!2!2–2×36−24(iv) s{A}=246. This time we start by permuting the 7 objects AA, B, B, C, C, D, D in7!2!3ways and then removing from it the subsets {A,B}, {A,C}, {A,D}, {A,B,C}, {A,B,D}, {A,C,D} and {A,B,C,D}. Hence s{A}=7!2!3−3×36−3×84–24Assembling the total number of perms with some adjacency:There are 4C4 subsets of type (i) ie {A,864,C,D}Total number of 8-perms of this type is 24.There are 4C3 subsets of type (ii) ie {A,864,C} {A,864,D} {A,C,D} {864,C,D}Total number of 8-perms of this type is 4×36=144There are 4C2 subsets of type (iii) ie {A,864} {A,C} {A,D} {864,C} {864,D} {C,D}Total number of 8-perms of this type is 6×84=504Finally there are 4C1 subsets of type (iv) ie {A} {864} {C} {D}Total number of 8-perms of this type=4×246=984Therefore, number of perms with some adjacency is  24+144+504+984= 
<gadget id="calculator"> 24+144+504+984</gadget>
<output>1_656</output>
1656Number of perms with no adjacency is 2520−1656= 864   Answer : 864
<result>B</result>
Bob is about to hang his 8 shirts in the wardrobe. He has four different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to one another?