1.Algebra Booster

5.42 Algebra Booster
= 14 C 12
– 3 ¥ 8 C 6
+ 3
= 91 – 84 + 3
= 10
Alternate method
The number of possible triplets to get 15 are
(5, 5, 5), (5, 6, 4) and (6, 6, 4)
Thus, the number of possible ways
3! 3!
= + 3! +
3! 2!
= 1 + 6 + 3
= 10
232. The number of possible ways
= Co-efficient of x 16 in (x 3 + x 4 + … + x 7 ) 4
= Co-efficient of x 4 in (1 + x + x 2 + x 3 + x 4 ) 4
Ê
5
1 - x ˆ
= Co-efficient of x 4 in Á
Ë 1 - x
˜
¯
= Co-efficient of x 4 in (1 – x 5 ) 4 ¥ (1 – x) –4
= Co-efficient of x 4 in (1 – x) –4
= Co-efficient of x 4 in (1 + 4 C 1
x + 5 C 2
x 2 + 6 C 3
x 3
5
+ 7 C 4
x 4 + … )
= 7 C 4
= 35
Alternate method
The number of possible quadruplets are
(3, 3, 3, 7), (3, 3, 5, 5), (3, 3, 4, 6), (3, 4, 4, 5) and (4, 4,
4, 4).
Hence, the number of possible ways
4! 4! 4! 4! 4!
= + + + +
3! 2!2! 2! 2! 2!
= 4 + 6 + 12 + 12 + 1
= 35
233. (i) The total number of positive solutions
= 20+4–1 C 4–1
= 23 C 3
23.22.21
=
6
= 1771
(ii) Here, x ≥ 1, y ≥ 1, z ≥ 1, w ≥ 1
fi x – 1 ≥ 0, y – 1 ≥ 0, z – 1 ≥ 0, w – 1 ≥ 0
Let p = x – 1, q = y – 1, r = z – 1, s = w – 1
Now, x + y + z + w = 20
fi p + q + r + s + 4 = 20
fi p + q + r + s = 16
Hence, the number of required solutions
= 16+4–1 C 4–1
= 19 C 3
19.18.17
=
6
= 969
234. Given 3x + y + z = 24, x ≥ 0, y ≥ 0, z ≥ 0
Let z = k.
Then y + z = 24 – 3k
So, 0 £ 24 – 3k £ 24
fi –24 £ –3k £ 24
fi 0 £ k £ 8
Thus, the number of integral solutions
= 24–3k+2–1 C 2–1
= 25–3k C 1
= (25 – 3k)
Hence, the number of integral solutions
8
= (25 -3 k)
Â
k = 0
8 8
Â
= 25 -3
Â
k= 0 k=
0
k
3¥ 8¥
9
= 25 ¥ 9 -
2
= 225 – 108
= 117
235. We have 2x + 2y + z = 10
10 - z
fi x+ y =
2
So, 0 £ x + y £ 10
Let x + y + t = 10, x ≥ 0, y ≥ 0, z ≥ 0
Hence, the number of non-negative integral solutions
= 20+4–1 C 4–1
= 23 C 3
12.11
= = 66
2
236. Here, x – 1 ≥ 0, y – 2 ≥ 0, z – 3 ≥ 0, t – 4 ≥ 0
Let p = x – 1, q = y – 2, r = z – 3, s = t – 4
fi x = p + 1, y = q + 2, z = r + 3, t = s + 4
Now, x + y + z + t = 29
fi p + q + r + s = 29 – 10 = 19
Hence, the number of non-negative solutions
= 19+4–1 C 4–1
= 22 C 3
22.21.20
=
6
= 1540
237. Given system of equations are
x + y + z – t + u = 20, x + y + z = 5
x + y + z = 5
…(i)
fi t + u = 15 …(ii)
Now, the number of non-negative integral solutions of
(i) = 5+3–1 C 3–1
= 7 C 2
= 21
Also, the number of non-negative integral solutions of
(ii)
= 15+2–1 C 2–1
= 16 C 1
= 16