The arrangement is as shown below.
where sn are the students and Bn are bamboos.
Total number of ways the 12 students can be arrayed in 12 different positions = 12!
Total number of ways the 3 bamboos can be arranged
=
3! /2!
(because two bamboos are red, hence identical)
= 3
Total number of possible arrangements
= 12! × 3 .
Seat 4 has to be occupied always.
Since no two persons are to be consecutive seating this can be done in 4 ways, i.e. (1, 4, 6), ( 1, 4, 7),
( 2, 4, 6),( 2, 4, 7).
3 seats can be chosen at random in 7C3 or 35 ways. ∴ p =
4/35
Total cases = 6 3 = 216. Favourable cases = 20. Ans 20/216 = 5/54. The cases are 123, 124, 125, 126, 134,
135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345, 346, 356, 456.
Total number of 5 lettered words that can be formed using 10 letters, with repetition = 105
total number of 5 lettered words that can be formed using 10 letters, such that the all letter used are
distinct = 10 × 9 × 8 × 7 × 6 = 30240.
Total words that have atleast one letter repeated
= 105 – 30240
= 69760.