Question

Q 2. find no of ways in which 4 particular persons a,b,c,d and 6 more persons can stand in a queue so that a always stand before

b. b always stand before c, and c always stand before

d.

Answer 1

Acc to the question there are 10 persons.

Now don't make a mistake by keeping abcd always together and considering no person can stand in between them.

So firstly, we can arrange this 4 person among 10 seats in 10C4 ways.

Now the remaining 6 person can be arranged again in 6! ways.

Therefore, no of ways in which 4 particular persons a,b,c,d and 6 more persons can stand in a queue so that a always stand before b, b always stand before c, and c always stand before d = 6! * 10C4 = 151200.

Answer 2

Answer:

5040

Step-by-step explanation:

a,b,c,d are grouped into one form i.e., 1 way

remaining 6 members stand in 6 ways

total=factorial (6+1) ways

        =7*6*5*4*3*2*1

=5040