Maths-
Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -
Maths-General
- 109
- 118
- 119
- None of these
Answer:The correct answer is: 119Required number of ways 
D(r)
=
= 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)
= 10 + 20 + 45 + 44
= 119.
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maths-
In how many ways can we get a sum of at most 17 by throwing six distinct dice -
x1 + x2 + x3 + x4 + x5 + x6 17
When 1
xi 6, i = 1, 2, 3, …..6
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
0 Required number of ways
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
When 1
xi 6, i = 1, 2, 3, …..6Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
0 Required number of ways= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
In how many ways can we get a sum of at most 17 by throwing six distinct dice -
maths-General
x1 + x2 + x3 + x4 + x5 + x6 17
When 1 xi 6, i = 1, 2, 3, …..6
Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7 0 Required number of ways
= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
When 1
xi 6, i = 1, 2, 3, …..6Let x7 be a variable such that
x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
Clearly x7
0 Required number of ways= Coefficient of x17 in (x1 + x2 + ….. + x6)6
(1 + x + x2 + …..)
= Coefficient of x11 in
= Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
= Coefficient of x11 in (1 – x)–7 – 6C1 × coefficient of x5 in (1 – x)–7
= 11+7–1C7–1 – 6C1 × 7+5–1C7–1
= 17C6 – 6 × 11C6 = 9604.
maths-
The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1
Let t = z + 1
Equation reduces to x + y + t = 25
1 x 5, 12 y 18, t 0
Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in
(1 – x)–1
= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
Equation reduces to x + y + t = 25
1
x 5, 12 y 18, t 0Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in
(1 – x)–1= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1
maths-General
Let t = z + 1
Equation reduces to x + y + t = 25
1 x 5, 12 y 18, t 0
Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in (1 – x)–1
= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
Equation reduces to x + y + t = 25
1
x 5, 12 y 18, t 0Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in
(1 – x)–1= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
maths-
For x 
R, let [x] denote the greatest integer x, then value of
+
+ 
+…+
is -
For 0 r 66, 0 
< 
– < – 0
–
– < – – –
= –1 for 0 r 66
Also, for 67 r 100, 
1
–1 – –
– – 1 – – – –
= –2 for 67 r 100
Hence,
= 67(–1) + 2(–34) = –135.
r 66, 0 < – < – 0– – < – – – = –1 for 0 r 66Also, for 67
r 100, 1–1 – – – – 1 – – – – = –2 for 67 r 100Hence,
= 67(–1) + 2(–34) = –135.For x R, let [x] denote the greatest integer x, then value of++ +…+is -
maths-General
For 0 r 66, 0 <
– < – 0
– – < – – –
= –1 for 0 r 66
Also, for 67 r 100, 1
–1 – –
– – 1 – – – –
= –2 for 67 r 100
Hence,= 67(–1) + 2(–34) = –135.
r 66, 0 < – < – 0– – < – – – = –1 for 0 r 66Also, for 67
r 100, 1–1 – – – – 1 – – – – = –2 for 67 r 100Hence,
= 67(–1) + 2(–34) = –135.maths-
The number of non negative integral solutions of equation 3x + y + z = 24
3x + y + z = 24, x 0, y 0, z 0
Let x = k y + z = 24 – 3k …(1)
24 – 3k 0 k 8
0 k 8
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
= 25 × 9 – 
= 225 – 108 = 117.
0, y 0, z 0Let x = k
y + z = 24 – 3k …(1) 24 – 3k 0 k 8 0 k 8For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
= 25 × 9 – = 225 – 108 = 117.The number of non negative integral solutions of equation 3x + y + z = 24
maths-General
3x + y + z = 24, x 0, y 0, z 0
Let x = k y + z = 24 – 3k …(1)
24 – 3k 0 k 8
0 k 8
For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
= 25 × 9 – = 225 – 108 = 117.
0, y 0, z 0Let x = k
y + z = 24 – 3k …(1) 24 – 3k 0 k 8 0 k 8For fixed value of k the number of solutions of (1) is
24–3k+2–1C2–1
= 25–3kC1
= 25 – 3k
Hence number of solutions
= 25 × 9 – = 225 – 108 = 117.maths-
If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -
We have (a3 – a) + (b3 – b) + (c3 – c)
= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
6 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + c).
= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
6 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + c).If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -
maths-General
We have (a3 – a) + (b3 – b) + (c3 – c)
= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
6 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + c).
= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)
Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,
6 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + c).maths-
Sum of divisors of 25 ·37 ·53 · 72 is –
Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0 l 5, 0 m 7, 0 n 3 and 0 p 2
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
.
l 5, 0 m 7, 0 n 3 and 0 p 2Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
.Sum of divisors of 25 ·37 ·53 · 72 is –
maths-General
Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0 l 5, 0 m 7, 0 n 3 and 0 p 2
Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=.
l 5, 0 m 7, 0 n 3 and 0 p 2Hence the sum of the divisors
= (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
=
=
.maths-
There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -
Let the number of yellow balls be x, that of black be 2x and that of green be y. Then
x + 2x + y = 20 or 3x + y = 20
y = 20 – 3x.
As 0 y 20, we get 0 20 – 3x 20
0 3x 20 or 0 x 6
The number of ways of selecting the balls is 7.
x + 2x + y = 20 or 3x + y = 20
y = 20 – 3x.As 0
y 20, we get 0 20 – 3x 20 0 3x 20 or 0 x 6The number of ways of selecting the balls is 7.There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -
maths-General
Let the number of yellow balls be x, that of black be 2x and that of green be y. Then
x + 2x + y = 20 or 3x + y = 20
y = 20 – 3x.
As 0 y 20, we get 0 20 – 3x 20
0 3x 20 or 0 x 6
The number of ways of selecting the balls is 7.
x + 2x + y = 20 or 3x + y = 20
y = 20 – 3x.As 0
y 20, we get 0 20 – 3x 20 0 3x 20 or 0 x 6The number of ways of selecting the balls is 7.maths-
A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -
Number of groups having 4 boys and 1 girl
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85 g = 5.
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85
g = 5.A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -
maths-General
Number of groups having 4 boys and 1 girl
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85 g = 5.
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85
g = 5.maths-
The length of the perpendicular from the pole to the straight line 
is
The length of the perpendicular from the pole to the straight line is
maths-General
maths-
The condition for the lines 
and 
to be perpendicular is
The condition for the lines and to be perpendicular is
maths-General
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If f : R →R; f(x) = sin x + x, then the value of 
(f-1 (x)) dx, is equal to
If f : R →R; f(x) = sin x + x, then the value of (f-1 (x)) dx, is equal to
maths-General
maths-
The polar equation of the straight line with intercepts 'a' and 'b' on the rays 
and 
respectively is
The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is
maths-General
maths-
The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is
The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is
maths-General
maths-
The polar equation of 
axy is
The polar equation of axy is
maths-General
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If x, y are rational number such that 
Then
If x, y are rational number such that Then
maths-General