Math, asked by kh8478535, 1 year ago

6 cubical boxes A, B, C, D, E and F are arranged one above the other vertically. The boxes are arranged subject to following conditions:
(i) The number of boxes above F is the same as the number of boxes below C.
(ii) E is the only box below B.
(iii) A is not the top-most box.
In how many ways can the boxes be arranges ?​

Answers

Answered by amitnrw
1

Answer:

2 ways

Step-by-step explanation:

2 ways

E is the only box below B.

=> E is at the Bottom most

& B is just above that

now maximum box above F can be 3

& minimum box below C can be

so two combinations possible

3 Boxes above F  & 3 Boxes below C

& 2 Boxes above F  &  2 Boxes below C

hence C & F  will take  (3rd & 4th Positions)

Now we left with two top position & two boxes A & D

but A  is not the top-most box.

=> D is at top most & A is just below that

Hence total 2 Ways :

D

A

C

F

B

E

D

A

F

C

B

E

Answered by nuuk
0

Answer:D A F C B E

Step-by-step explanation:

Given

Six Boxes Placed one over other

It is given no of boxes above F and below C is same i.e. they should ideally be at 3 rd and 4 th position from top.

E is the only box below B i.e. E is at the bottom .

A is not the top most therefore it should be at 2 nd Position.

Therefore D is at the top

Thus the order is D A F C B E

Similar questions