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#### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

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^{K + 6}C_{6} – 6 ^{K – 4}C_{6}
^{K}C_{6} – 6 ^{K – 4}C_{6}
^{K}C_{6} – 6 ^{K}C_{4}
^{(K + 6)}C_{6} – ^{(K – 4)}C_{6}

^{K + 6}C_{6}– 6^{K – 4}C_{6}^{K}C_{6}– 6^{K – 4}C_{6}^{K}C_{6}– 6^{K}C_{4}^{(K + 6)}C_{6}–^{(K – 4)}C_{6}#### Answer:The correct answer is: ^{K + 6}C_{6} – 6 ^{K – 4}C_{6}The required no. of ways = no. of solution of the equation (x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = K)

Where 0 x_{i} 9, i = 1, 2, …6, where 0 < K < 18

= Coefficient of x^{K} in (1 + x + x^{2} +….. + x^{9})^{6}

= Coefficient of x^{K} in

= Coefficient of x^{k} in (1 – 6x^{10} + 15 x^{20} – ….)

(1 + 6 C_{1}x + 7 C_{2} x^{2} + …. +(7 – K – 10 – 1) C_{K–10} x^{K–10} + ….+(7 + K – 1) C_{K }x^{K} + …)

= ^{k + 6}C_{K} – 6. ^{K–4}C_{K–10}

= ^{k + 6}C_{6} – 6. ^{K–4}C_{6 }.

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### Related Questions to study

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#### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.#### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

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Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.maths-

#### If the line is a normal to the hyperbola then

#### If the line is a normal to the hyperbola then

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maths-

#### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

#### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

maths-General

maths-

#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

maths-General

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).maths-

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

maths-General

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

maths-

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

maths-General

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

maths-

#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

maths-General

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

maths-

#### Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{i}( 3) be the number of blanks between i^{th}and (i + 1)^{th}letters. Then,x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15 …..(1)The number of solutions of (1)

= coefficient of t

^{15}in (t^{3}+ t^{4}+….)^{4}= coefficient of t

^{3}in (1 – t)^{–4}= coefficient of t

^{3}in [1 +^{4}C_{1}+^{5}C_{2}t^{2}+^{6}C_{3 }t^{3}+ …..]=

^{6}C_{3}= 20.But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

#### Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -

maths-General

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{i}( 3) be the number of blanks between i^{th}and (i + 1)^{th}letters. Then,x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15 …..(1)The number of solutions of (1)

= coefficient of t

^{15}in (t^{3}+ t^{4}+….)^{4}= coefficient of t

^{3}in (1 – t)^{–4}= coefficient of t

^{3}in [1 +^{4}C_{1}+^{5}C_{2}t^{2}+^{6}C_{3 }t^{3}+ …..]=

^{6}C_{3}= 20.But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

maths-

#### The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

#### The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

maths-General

maths-

#### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

#### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

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maths-

#### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

#### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

maths-General

maths-

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

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maths-

The coefficient of in is

The coefficient of in is

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#### How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which not two S are adjacent ?

#### How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which not two S are adjacent ?

maths-General

maths-

#### The value of ^{50}C_{4} + is -

#### The value of ^{50}C_{4} + is -

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