(Q). In how many ways 15 students can be arranged in a row such that A is always ahead of B ?
Sol:- Without any condition 15 students can be arranged in 15! ways.
In these arrangements in 50% cases A will be ahead of B and in remaining 50% cases B will be ahead of A.
Hence number of ways when A is ahead of B =
(Q). In how many ways batting order of 11 players can be made out of 15 players
(i). If A & B are always rejected
(ii). If A & B are always rejected but C & D are always selected.
(iii). If two players can only play as wicket keeper.
Sol:-
(i). ∵ A & B are always rejected.
So we have to select 11 players out of remaining 15 – 2 = 13 players which can be done in ¹³C₁₁ ways
Now these 11 players can be arranged in 11! ways.
Hence total number of ways
= ¹³C₁₁.11! Answer
(ii). A & B are always rejected then we have to select 11 out of 15 – 2 = 13
again C & D are always selected then we have to select 11 – 2 = 9 players out of 13 – 2 = 11 players which can be done in ¹¹C₉ ways.
Now these selected 11 players can be arranged in 11! ways.
Hence total number of ways
= ¹¹C₉.11! Answer
(iii). From 2 wicket keeper only 1 is selected and this can be done in ²C₁ ways.
Now remaining 10 players can be selected out of 15 -2 = 13 players in ¹³C₁₀ ways.
So total number of ways of selection = ²C₁×¹³C₁₀
Now these 11 players can be arranged in 11! ways
Hence total number of ways = (²C₁×¹³C₁₀)(11!) Answer
(Q). If x is the number of ways in which 5 students can be arranged on 15 chairs in a straight line and y is the number of ways in which 5 students out of 15 students can be arranged on 5 chairs in a straight line. Then what is the relation between x & y ?
Sol:- x = (number of ways to arrange 5 students on 15 chairs in a straight line)
y = (number of ways to arrange 5 students out of 15 students in 5 chairs)
Hence x = y Answer