1.Algebra Booster

Permutations and Combinations 5.39
Q, R, S P
P, R, S Q
P, Q, S R
P, Q, R S
Hence, the number of possible ways
= 4 C 3
¥ 1 C 1
¥ 2!
= 4 ¥ 2 = 8
206. Set I Set II No. of ways
P, Q R, S
P, R Q, S
4 1
C
¥
C
3 1
2!
P, S Q, R
Q, R P, S
Q, S P, R
R, S P, Q
Hence, the number of possible ways
=
4 1
C3¥
C1
2!
= 3
207. Person I Person II No. of ways
P, Q R, S
P, R Q, S
P, S Q, R
Q, R P, S 4
C 2
¥ 2 C 2
Q, S P, R
R, S P, Q
Hence, the number of possible ways
= 4 C 2
¥ 2 C 2
= 6
208. The possible triplets of 5 are
(1, 1, 3) and (1, 2, 2).
Hence, the number of possible ways
3! 3!
= 5! ¥ + 5! ¥
2! 2!
= 120 ¥ 3 + 120 ¥ 3
= 360 + 360
= 720
Alternate method
The number of possible ways
= 5–1 C 3–1
¥ 5!
= 4 C 2
¥ 5!
= 6 ¥ 120
= 720
209. The possible triplets of 5 are
(1, 1, 3) and (1, 2, 2).
Hence, the number of possible ways
5! 5!
= ¥ 3! + ¥ 3!
1!1!3! 2! 2!1!
= 120 + 180
= 300
210. The possible doublets of 4 are (1, 3) and (2, 2).
Hence, the number of possible ways
4! 4! 2!
= ¥ 2! + ¥
1!3! 2!2! 2!
4! 4!
= +
3 2!2!
24 24
= +
3 4
= 8+
6
= 14
211. The possible triplets of 5 are (1, 1, 3) and (1, 2, 2).
Hence, the number of possible ways
5! 3! 5! 3!
= ¥ + ¥
1!1!3! 2! 2! 2!1! 2!
5! 5! ¥ 3!
= +
2! 2!2!2!
= 60 + 90
= 150
212. The possible triplets are
(1, 2, 3), (1, 1, 4) and (2, 2, 2).
Hence, the number of possible ways
6! 6! 3! 6! 3!
= ¥ 3! + ¥ + ¥
1!2!3! 1!1!4! 2! 2!2!2! 3!
6! 6! 6!
= + +
2! 4¥
2! 2!2!2!
720 720 720
= + +
2 8 8
= 360 + 90 + 90
= 540
213. The number of possible ways
= 10–1 C 3–1
= 9 C 2
9¥
8
=
2
= 36
Alternate method
Here, the possible triplets are
(1, 3, 6), (1, 4, 5), (1, 2, 7), (2, 3, 5), (1, 1, 8), (2, 2, 6),
(2, 4, 4) and (3, 3, 4)
Hence, the number of possible ways
= (1 ¥ 3! + 1 ¥ 3! + 1 ¥ 3! + 1 ¥ 3!)
Ê 3! 3! 3! 3! ˆ
+ Á1¥ + 1¥ + 1¥ + 1¥
Ë 2! 2! 2! 2! ˜
¯
= 6 ¥ 4 + 3 ¥ 4
= 24 + 12
= 36