(Q). In how many ways 5 distinct volumes of chemistry & 7 distinct volume of botany books can be arranged on a bookshelf such that no two botany books are together ?
Sol:-
•C₁•C₂•C₃•C₄•C₅•
From above figure we have 6 places (marked by •) to place 7 botany books.
So there is no arrangement possible to place botany books such that no two botany books are together.
Hence No arrangement possible Answer
(Q). In how many ways 15 students can be arranged in a row such that x is ahead of y and who in turn is ahead of z ?
Sol:- we can arrange three students x, y & z in 3! = 6 ways in which only one way ⟹ xyz i.e. when x is ahead of y and who in turn is ahead of z is of our use.
Now without any restriction 15 students can be arranged in 15! ways.
& we know that every 6 only 1 case satisfy the given condition
Hence total number of ways = Answer
Method (2): _x_y_z_
we have selected x, y & z and placed them in the desired condition.
Now there are 4 spaces left for next student.
So Fourth student can be placed at these 4 places in 4 ways & this will increase one more space for next student to be placed.
∴ 5ᵗʰ student can go at 5 empty places in 5 ways & after placing this student one more space will be created for next student & this process will go on till all student ae placed.
.
.
.
.
.
15ᵗʰ student can be placed in 15 ways.
Hence total number of ways to get desired answer
= 4×5×6×7×………….×15
=
= Answer
(Q). What is the total number of signals that can be made by using 6 flags of different colour when any number of them may be used ?
Sol:- The question can be solved by making different cases:-
case(i): when all the 6 flags are used:-
then number of signal = number of ways to arrange = 6!
case(ii): when 5 flags are used:-
Number of ways to select 5 flags out of 6 flags = ⁶C₅
Now these 5 flags can be arranged in 5! ways
So number of signal in this case = ⁶C₅.5!
case(iii): when 4 flags are used:-
Number of signal in this case = ⁶C₄.4!
case(iv): when 3 flags are used:-
Number of signal in this case = ⁶C₃.3!
case(v): when 2 flags are used:-
Number of signal in this case = ⁶C₂.2!
case(vi): when 1 flag is used:-
Number of signal in this case = ⁶C₁.1!
Hence total number of signals
= 6! + ⁶C₅.5! + ⁶C₄.4! + ⁶C₃.3! + ⁶C₂.2! + ⁶C₁.1!
= 1956 Answer