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Akshath K.
Research Consultant at WorldQuant LLC
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Basic Math
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Question:

How many odd numbers, greater than 600000 can be formed from digits 5, 6, 7, 8, 9, 0 if a.) Repetition is allowed. b.) Repetition is not allowed.

Akshath K.
Answer:

This question is from the topic of permutations and combinations and can be easily solved by analyzing the question. Let's take the first part where repetition is allowed. Given Digits: 5, 6, 7, 8, 9, 0 REPETITION ALLOWED a.) The first place can be filled by 6, 7, 8, 9 i.e in 4 ways to make the number greater than 600000. The last place can be filled by 5, 7, 9 i.e 3 ways as the number is to be odd and because of repetition. Hence, the number of ways of filling first and last place = 4x3 = 12 ways We have to fill in the remaining 4 places of the six-digit number i.e 2nd, 3rd, 4th and 5th place. Since repetition is allowed each place can be filled in 6 ways. Hence, the 4 places can be filled in = 6x6x6x6 = 1296 ways Thus by the fundamental theorem of permutations and combinations the total number of ways = Total numbers that can be formed = 1296 x 12 = 15552 Now, let's discuss the case if repetition is not allowed. REPETITION NOT ALLOWED b.) Discussing every possible case First Place Last place (Number to be odd) 6 5, 7, 9 = 3 ways 8 5, 7, 9 = 3 ways 7 5. 9 = 2 ways 9 5, 7 = 2 ways Total number of ways of filling the first and the last place under given condition is = 3 + 3 + 2 + 2 = 10 ways (It was 12 if repetition was allowed, see the difference) Having filled the first and the last places in 10 ways, we can fill in the remaining 4 places out of 4 digits (repetition not allowed) in 4! = 24 ways Thus by the fundamental theorem of permutations and combinations the total number of ways = Total numbers that can be formed = 10 x 24 = 240

C++ Programming
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Question:

Given an array of N elements, find the length of the longest subarray such that sum of the subarray is even. Input/ Output Examples Input : Size = 6, arr[] = {1, 2, 3, 2, 1, 4} Output : 5 Explanation: In the example the subarray in range [2, 6] has sum 12 which is even, so the length is 5. Input : Size = 4, arr[] = {1, 2, 3, 2} Output : 4

Akshath K.
Answer:

APPROACH : First, check if the total sum of the array is even. If the total sum of the array is even then the answer will be N. If the total sum of the array is not even, means it is ODD. So, the idea is to find an odd element from the array such that excluding that element and comparing the length of both parts of the array we can obtain the max length of the subarray with even sum. It is obvious that the subarray with even sum will exist in range [1, x) or (x, N], where 1 <= x <= N, and arr[x] is ODD. C++ CODE : #include <iostream> using namespace std; // Function to find length of the longest subarray such that sum of the subarray is even int maxLength(int a[], int n) { int sum = 0, len = 0; // Check if sum of complete array is even for (int i = 0; i < n; i++) sum += a[i]; if (sum % 2 == 0) // total sum is already even return n; // Find an index i such the a[i] is odd and compare length of both halfs excluding a[i] to // find max length subarray for (int i = 0; i < n; i++) { if (a[i] % 2 == 1) len = max(len, max(n - i - 1, i)); } return len; } // Driver Code int main() { int size; //Initializing size of array cin >> size; int a[size]; //Initializing size of array for (int i=0; i<size; i++) // Taking input for the array { cin >> a[i]; } cout << maxLength(a, n) << endl; // Printing the max length using our function return 0; }

Physics
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Question:

A billiard ball, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance h above the center line. The ball leaves the cue with a speed of u and because of its backward slipping motion eventually acquires a final speed of 9u/7. Show that h=4R/5 where R is the radius of the ball.

Akshath K.
Answer:

It's a very good question as it touches several different concepts of mechanics like rolling motion, impulse equations and conservation of momentum. Using these concepts we'll properly solve this problem stepwise. We have, F = Forward force applied by cue on the ball T = Torque applied by cue on the ball = (Force)*(Perpendicular distance from the axis of rotation) = F*h R = Radius of ball m = Mass of ball I = Moment of inertia of ball (solid sphere) = (2m(R^2))/5 u = Initial linear speed just after it leaves the cue v = Final speed after slipping is ceased h = Height above the center the ball should be hit w1 = Initial angular velocity just after it leaves the cue w2 = Final angular velocity after slipping is ceased Now, the maximum friction acts in forward direction till the slipping continues. We know, v = R*w2 => w2 = v/R (For Rolling motion constraints) Given, v = 9u/7 ____________________(i) => w2 = (9*u)/(7R) __________________ (ii) Applying the impulse equations, (Linear Impulse = Change in linear momentum) F.dt = mu - 0 = mu ___________________(iii) (Angular Impulse = Change in angular momentum) T.dt = I*(w1) Fh.dt = (2m(R^2)w1)/5 ___________________(iv) Angular momentum about the bottommost point will remain conserved since there is no net external torque applied at that point (Initial Angular Momentum = Final Angular Momentum) I*w1 + mRu = I*w2 + mRv (2m(R^2)w1)/5 + mRu = ((2m(R^2))/5)*(9u)/(7R)) + 9mRv/7 __________(v) Now after solving equations (iii), (iv) and (v) simultaneously, we get h = 4R/5

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