1.Algebra Booster

Permutations and Combinations 5.41
Hence, the number of possible ways
= 35 + 20 – 4 – 8
= 5! ¥ 4
= 43
= 1320 ¥ 4
The number of permutations of 4 letters is = Co-efficient
of x 4 in
= 5280
225. 2 letters can be selected out of 5 letters in 5 2 2
C 2
ways. For
Ê
2 3 2
x x x ˆ Ê x x ˆ
each such selections, the remaining 3 letters placed in 4! ¥ Á1 + + + ¥ 1 + + ¥ (1 + x)
Ë 1! 2! 3!
˜
¯
Á
Ë 1! 2!
˜
¯
wrong envelopes in D 3
ways.
Hence, the number of possible selections
= Co-efficient of x 4 in
= 5 C 2
¥ D
2 2
3
Ê
2 3 2
x x x ˆ Ê x x ˆ
= 10 ¥ 2
4! ¥ Á1 + + + ¥ 1 + + ¥ (1 + x)
Ë
˜
1 2 6 ¯
Á
Ë
˜
1 2 ¯
= 20
226. The number of ways to place at least 2 balls in boxes of = Co-efficient of x 4 in
the same color
Ê
2 4 3 7 4ˆ
4! ¥
= The number of ways without restriction – the
Á1 + 2x + 2x + x + x
Ë
˜
3 12 ¯
number of ways all 5 balls are wrongly placed – the
Ê
2 3 5 4ˆ
number of ways to select a ball in right box ¥ number
¥ Á1+ 3x + 4x + 3x + x
Ë
˜
4 ¯
of ways in which remaining 4 balls in 4 wrong boxes
= 5! – 44 – 5 C 1
¥ D Ê 7 5
ˆ
4
= 4! ¥ Á + + (6 + 4) + 8
= 120 – 44 – 5 ¥ 9
Ë
˜
12 4
¯
= 120 – 44 – 45
22
= 24 ¥ + 24 ¥ 18
= 120 – 89
12
= 31
= 44 + 432
227. The number of possible ways
= 476
= 0 letter in wrong envelope and 6 are correct 230. There are 11 letters in which 3 Es. 3 Ns, 2 Ds, 1 I, 1 P
+ 1 letter in wrong envelope and 5 are correct and 1 T, respectively.
+ 2 letters in wrong envelopes and 4 are correct Hence, the total number of selections
+ 3 letters in wrong envelopes and 3 are correct
= Co-efficient of x 5 in (1 + x + x 2 + x 3 ) 2
= 1 + 0 (not possible) + 6 C 4
¥ D 2
+ 6 C 3
¥ D 3
¥ (1 + x + x 2 ) ¥ (1 + x) 3
6¥ 5 6¥ 5¥
4
= Co-efficient of x
= 1+ ¥ 1+ ¥ 2
5 in
2 6
2
Ê
4 3
1- x ˆ Ê1-
x ˆ
3
= 1 + 15 + 40
Á ¥ ¥ (1 + x)
Ë 1- x
˜
¯
Á
Ë 1-
x
˜
¯
= 56
228. The number of possible ways = the number of ways
= Co-efficient of x 5 in
pick up 1 right coat and 5 wrong coats
(1 – 2x 4 )(1 – x 3 ) ¥ (1 – x) –3 ¥ (1 + x) 3
= 6 C 1
¥ D 5
= Co-efficient of x 5 in
= 6 ¥ 44
(1 – x 3 – 2x 4 ) ¥ (1 + 3x + 3x 2 + x 3 ) ¥ (1 – x) –3
= 264
= Co-efficient of x
229. There are 11 letters in which 3 Es, 3 Ns, 2 Gs, 2 Is and
5 in
1 R.
(1 + 3x + 3x 2 – 5x 4 – 9x 5 )
= Co-efficient of x 4 in
(1 + 3 C 1
x + 4 C 2
x 2 + 5 C 3
x 3 + 6 C 4
x 4 + 7 C 5
x 5 )
(1 + x + x 2 + x 3 ) 2 ¥ (1 + x + x 2 ) ¥ (1 + x) 3
= ( 7 C 5
– 9) + (3 ¥ 6 C 4
– 15) + (3 ¥ 5 C 3
)
= Co-efficient of x 4 in
= 21 – 9 + 45 – 15 + 30
4
2
3
2
Ê1- x ˆ Ê1-
x ˆ
= 12 + 30 + 30
Á ¥ ¥ (1 + x)
Ë 1- x
˜
¯
Á
Ë 1-
x
˜
¯
= 72
231. The number of possible ways
= Co-efficient of x 4 in
= Co-efficient of x 15 in (x + x 2 + … + x 6 ) 3
{(1 – x 4 )(1 – x 3 )} 2 (1 + x)(1 – x) –4
= Co-efficient of x 12 in (x + … + x 5 ) 3
= Co-efficient of x 4 in
3
(1 – x 3 – x 4 )(1 + x)(1 – x) –4
Ê
6
1 - x ˆ
= Co-efficient of x 12 in Á
= Co-efficient of x 4 in
Ë 1 - x
˜
¯
(1 – x – 2x 3 – 4x 4 ) (1 + 4 C 1
x + 5 C 2
x 2 + 6 C 3
x 3
= Co-efficient of x 12 in (1 – x 6 ) 3 ¥ (1 – x) –3
+ 7 C 4
x 4 )
= Co-efficient of x 12 in (1 – 3x 6 + 3x 12 )
= 7 C 4
+ 6 C 3
– 4 – 4. 2 C 1 (1 + 3 C 1
x + 4 C 2
x 2 + … + 8 C 6
x 6 + … + 14 C 12
x 12 )