Permutations with restrictions

Question

$Z =\left \{ {A,B,C,D,E,F,G} \right \}$

How many permutations of all elements of set $Z$ are possible when 

1. A cannot appear after D and C cannot appear after F
2. C cannot appear after D 

Comments

thanks sourav. and Balaji Jegan

---

@balaji jegan  please explain your method?? im not getting answer for 1st questtion

---

Totally 7 distinct letters

so 7! ways

now, A can't appear after D How any ways can we arrange A and D?

2! ways

AD should be the correct order not DA

so we divide 7!/2!

same thing can be said for CF

Answer 1

1. $ \overline{A} \cap \overline{B} = U - (A \cup B)$

$ A \cup B  $   $\text{ means that ---- A can appear after D or C can appear after F}$

$\text{No of permutations when A can appear after D = } $ $6! $

$\text{No of permutations when C can appear after F = } $ $6! $

$\text{No of permutations when A can appear after D  and C can appear after F = } $ $5! $

$\text{Total no of permutations for Z = 7!}$

$ 7! - (6! + 6! -5!) = 3720$

 

2.  $\text{Similarly }$ $7! - 6! = 4320$

Comments

$\text{Consider that A can appear after D as DA. Means now we have 6 symbols so there will be 6!}$ $\text{permutations. Similarly for FC. Now if both the conditions are together then we just need }$ $\text{to subtract it from entire permutation}$

---

!KARAN no you misinterpreted the question, however the answer is in the comments

---

i got the same answer for part 2(2520)

Answer 2

 

subpart 2