The numbers of integers between 1 and 106 have the sum of their-Turito

Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

The numbers of integers between 1 and 10⁶ have the sum of their digit equal to K(where 0 < K < 18) is -

^{(K + 6)}C₆ – ^{(K – 4)}C₆
^KC₆ – 6 ^KC₄
^KC₆ – 6 ^{K – 4}C₆
^{K + 6}C₆ – 6 ^{K – 4}C₆

The correct answer is: ^{K + 6}C₆ – 6 ^{K – 4}C₆

The required no. of ways = no. of solution of the equation (x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = K)
Where 0  x_i 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of x^K in (1 + x + x² +….. + x⁹)⁶
= Coefficient of x^K in $open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent$
= Coefficient of x^k in (1 – 6x¹⁰ + 15 x²⁰ – ….)
(1 + 6 C₁x + 7 C₂ x² + …. +(7 – K – 10 – 1) C_K–10 x^K–10 + ….+(7 + K – 1) C_Kx^K + …)
= ^{k + 6}C_K – 6. ^K–4C_K–10
= ^{k + 6}C₆ – 6. ^K–4C₆.

Book A Free Demo

Name*

Mobile*

+91

Email*

Grade*

I agree to get WhatsApp notifications & Marketing updates

Related Questions to study

The straight lines I₁, I₂, I₃ are parallel and lie in the same plane. A total number of m points are taken on I₁ ; n points on I₂, k points on I₃. 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}C₃. But out of these m + n + k points, m points lie on I₁, n points lie on I₂ and k points lie on I₃ and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
^{m + n +} ^kC₃ – ^mC₃ – ⁿC₃ – ^kC₃.

The straight lines I₁, I₂, I₃ are parallel and lie in the same plane. A total number of m points are taken on I₁ ; n points on I₂, k points on I₃. 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}C₃. But out of these m + n + k points, m points lie on I₁, n points lie on I₂ and k points lie on I₃ and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
^{m + n +} ^kC₃ – ^mC₃ – ⁿC₃ – ^kC₃.

If the line $l x plus m y equals 1$ is a normal to the hyperbola $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ then $fraction numerator a to the power of 2 end exponent over denominator l to the power of 2 end exponent end fraction minus fraction numerator b to the power of 2 end exponent over denominator m to the power of 2 end exponent end fraction equals$

If the line $l x plus m y equals 1$ is a normal to the hyperbola $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ then $fraction numerator a to the power of 2 end exponent over denominator l to the power of 2 end exponent end fraction minus fraction numerator b to the power of 2 end exponent over denominator m to the power of 2 end exponent end fraction equals$

If the tangents drawn from a point on the hyperbola $x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent$ to the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ make angles α and β with the transverse axis of the hyperbola, then

A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.
We have given: the tangents drawn from a point on the hyperbola

x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent

fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1

make angles α and β with the transverse axis of the hyperbola.

N o w space l e t s space t a k e space a space p o i n t space P space o n space t h e space h y p e r b o l a space w i t h space c o o r d i n a t e s space left parenthesis h comma k right parenthesis comma space t h e n colon h squared minus k squared equals a squared minus b squared N o w space t h e space e q u a t i o n space o f space tan g e n t space t o space t h e space e l l i p s e space x squared over a squared plus y squared over b squared equals 1 space w i l l space b e colon y equals m x plus square root of a squared m squared plus b squared end root S i m p l i f y i n g space i t comma space w e space g e t colon left parenthesis k minus m h right parenthesis squared equals a to the power of 2 end exponent m to the power of 2 end exponent plus b squared m squared left parenthesis h squared minus a squared right parenthesis minus 2 m h k plus k squared minus b squared equals 0 N o w space l e t space m 1 space a n d space m 2 space b e space t h e space r o o t space o f space t h e space e q u a t i o n space f o r m e d comma space t h e n space w e space g e t colon m 1 m 2 equals fraction numerator k squared minus b squared over denominator h squared minus a squared end fraction equals 1 N o w space u sin g space e q u a t i o n space 1 comma space w e space g e t colon tan alpha space tan beta equals 1

If the tangents drawn from a point on the hyperbola $x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent$ to the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ make angles α and β with the transverse axis of the hyperbola, then

A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.
We have given: the tangents drawn from a point on the hyperbola

x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent

fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1

make angles α and β with the transverse axis of the hyperbola.

N o w space l e t s space t a k e space a space p o i n t space P space o n space t h e space h y p e r b o l a space w i t h space c o o r d i n a t e s space left parenthesis h comma k right parenthesis comma space t h e n colon h squared minus k squared equals a squared minus b squared N o w space t h e space e q u a t i o n space o f space tan g e n t space t o space t h e space e l l i p s e space x squared over a squared plus y squared over b squared equals 1 space w i l l space b e colon y equals m x plus square root of a squared m squared plus b squared end root S i m p l i f y i n g space i t comma space w e space g e t colon left parenthesis k minus m h right parenthesis squared equals a to the power of 2 end exponent m to the power of 2 end exponent plus b squared m squared left parenthesis h squared minus a squared right parenthesis minus 2 m h k plus k squared minus b squared equals 0 N o w space l e t space m 1 space a n d space m 2 space b e space t h e space r o o t space o f space t h e space e q u a t i o n space f o r m e d comma space t h e n space w e space g e t colon m 1 m 2 equals fraction numerator k squared minus b squared over denominator h squared minus a squared end fraction equals 1 N o w space u sin g space e q u a t i o n space 1 comma space w e space g e t colon tan alpha space tan beta equals 1

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

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r²t⁴s², then the number of ordered pair (p, q) is –

An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.
Now we have given the LCM as: r²t⁴s²
Consider following cases:
Case 1: if p contains r² then q will have r^k, for the value k=0,1.
So 2 ways.
Case 2: if q contains r² then p will have r^k, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r²
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r²t⁴s², then the number of ordered pair (p, q) is –

An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.
Now we have given the LCM as: r²t⁴s²
Consider following cases:
Case 1: if p contains r² then q will have r^k, for the value k=0,1.
So 2 ways.
Case 2: if q contains r² then p will have r^k, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r²
So 1way.
So after this we can say that:
exponent of r=2+2+1 = 5 ways.
Similarly
exponent of t=4+4+1=9 ways.
exponent of s=2+2+1 = 5 ways.
So total ways will be:
5 x 5 x 9 = 225 ways.

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

Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m²n²
So it is m²n².

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

Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
Then the number of rectangles will be:
(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m²n²
So it is m²n².

ⁿC_r + ²ⁿC_r+1 + ⁿC^r+2 is equal to (2 $less or equal than$ r $less or equal than$ n)

ⁿC_r + ²ⁿC_r+1 + ⁿC^r+2 is equal to (2 $less or equal than$ r $less or equal than$ n)

The coefficient of $x to the power of n$ in $fraction numerator x plus 1 over denominator left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis end fraction$ is

The coefficient of $x to the power of n$ in $fraction numerator x plus 1 over denominator left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis end fraction$ is

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 ?

The value of ⁵⁰C₄ + $not stretchy sum subscript r equals 1 end subscript superscript 6 end superscript 56 minus r C subscript 3 end subscript$ is -

The value of ⁵⁰C₄ + $not stretchy sum subscript r equals 1 end subscript superscript 6 end superscript 56 minus r C subscript 3 end subscript$ is -

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is-

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is-

A mass of $100 blank g$ strikes the wall with speed $5 blank m divided by s$ at an angle as shown in figure and it rebounds with the same speed. If the contact time is $2 cross times 10 to the power of negative 3 end exponent s e c$ , what is the force applied on the mass by the wall

Force = Rate of change of momentum
Initial momentum

stack P with rightwards arrow on top subscript 1 end subscript equals m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top

stack P with rightwards arrow on top subscript 2 end subscript equals negative m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top

therefore stack F with rightwards arrow on top equals fraction numerator increment stack P with rightwards arrow on top over denominator increment t end fraction equals fraction numerator negative 2 m v sin invisible function application theta over denominator 2 cross times 10 to the power of negative 3 end exponent end fraction

m equals 0.1 blank k g comma blank v equals 5 blank m divided by s comma blank theta equals 60 degree

Force on the ball

stack F with rightwards arrow on top equals negative 250 square root of 3 N

Negative sign indicates direction of the force

A mass of $100 blank g$ strikes the wall with speed $5 blank m divided by s$ at an angle as shown in figure and it rebounds with the same speed. If the contact time is $2 cross times 10 to the power of negative 3 end exponent s e c$ , what is the force applied on the mass by the wall

physics-General

Force = Rate of change of momentum
Initial momentum

stack P with rightwards arrow on top subscript 1 end subscript equals m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top

stack P with rightwards arrow on top subscript 2 end subscript equals negative m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top

therefore stack F with rightwards arrow on top equals fraction numerator increment stack P with rightwards arrow on top over denominator increment t end fraction equals fraction numerator negative 2 m v sin invisible function application theta over denominator 2 cross times 10 to the power of negative 3 end exponent end fraction

m equals 0.1 blank k g comma blank v equals 5 blank m divided by s comma blank theta equals 60 degree

Force on the ball

stack F with rightwards arrow on top equals negative 250 square root of 3 N

Negative sign indicates direction of the force

If the locus of the mid points of the chords of the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ , drawn parallel to $y equals m subscript 1 end subscript x$ is $y equals m subscript 2 end subscript x$ then $m subscript 1 end subscript m subscript 2 end subscript equals$

If the locus of the mid points of the chords of the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ , drawn parallel to $y equals m subscript 1 end subscript x$ is $y equals m subscript 2 end subscript x$ then $m subscript 1 end subscript m subscript 2 end subscript equals$

If ⁿC_r denotes the number of combinations of n things taken r at a time, then the expression ⁿC_r+1 + ⁿC_{r –1} + 2 × ⁿC_r equals-

If ⁿC_r denotes the number of combinations of n things taken r at a time, then the expression ⁿC_r+1 + ⁿC_{r –1} + 2 × ⁿC_r equals-

The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question is -

The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question is -