C is the daughter-in-law of A who is the grandmother of F means C is the mother of F. But F is the son of B. 

So. B is C's husband. But C, the lawyer is married to the Doctor So. B is the Doctor. F, the Accountant  will be the son of B and C. E is the unmarried Engineer 

So. the other mamed couple can be that of grandmother of F i.e, A and D.  But D, the Salesman, is married to the Lady Teacher. 

So. D, the Salesman, is the grandfather of F. father of B and the husband of A, the Lady Teacher. 

Clearly, from the given data, the relation between E and F cannot be determined. 

D is father of A and grandfather of F. So, A is father of F. 

Thus. D and A are the two fathers. C is the sister of F So. C is the daughter of A. 

Since there is only one mother, it is evident that E is the wife of A and hence the mother of C and F. 

So, B is brother of A There are three brothers. So. F is the brother of C.

A is E's husband

E is the daughter of A and F is the mother of E So, A is the father of E and hence the husband of F. Now. 

D is the daughter of F So. D and E are the daughters of A and F Also. A is the son of C. 

Now, only B remains Thus. B and C are the parents of A. 

Clearly, the females in the family are : either B or C, F, D and E. 

C is wife of B. So, one couple is BC. Now, D in grandmother of A, B is the son of D and his wife C is the mother of F. SO, D is also the grandmother of F But F is the granddaughter of E. 

So, E is the grandfather of F and the husband of D. Thus, DF is another couple 

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is 6C0=1 

The number of outcomes in which 1 coin turns head is =6C1=6 

The number of outcomes in which 2 coins turn heads is6C2=15 

The number of outcomes in which 3 coins turn heads is6C3=20

Therefore, total number of outcomes =1+6+15+20= 42 outcomes

He can arrange his schedule in 8P6 = 20160 ways

Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

 C is the daughter-in-law of A who is the grandmother of' F means C is the mother of F But F is the son of B. 

So. B is C's husband. But C, the lawyer. is married to the Doctor So. B is the Doctor. F, the Accountant, will be the non of B and C, E is the unmarried Engineer 

So, the other mamed couple can be that of grandmother of Fie A and 0 But D. the Salesman, is married to the Lady Teacher. 

So. D, the Salesman, is the grandfather of F, father of B and the husband of A, the Lady Teacher. 

D is the grandfather of F.