How many solutions are there of the equation a+b+c=12, where a,b,c are non-negative integers?

We can think of any solution of this equation as a string of twelve 1's and two 0's. For example, one such solution could be a=6, b=2 and c=4. In this case we would write the solution as 11111101101111. Written in this way, all solutions of the equation must consist of fourteen symbols; exactly twelve 1's and two 0's. The only difference between solutions is the choice of where to put the two 0's. Thus the problem reduces to finding out how many different positions the two 0's can take amongst the 1's. We know from the Binomial Theorem that the number of ways of choosing r objects from n objects is nCr = n!/r!(n-r)! In our case the 'objects' are the positions of the 1's and 0's and we want to know how many ways there are of choosing two positions from the fourteen available positions. Hence we can substitute into the above formula with n=14 and r=2 to find that the total number of solutions is 14C2=14!/(2!*12!)=91.

BM
Answered by Brendan M. Maths tutor

5954 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

integrate cos(2x) + sin(3x)


Find dy/dx when y = x(4x + 1)^1/2


Sketch 20x--x^2-2x^3


In the triangle ABC, AB = 16 cm, AC = 13 cm, angle ABC = 50 and angle BCA= x Find the two possible values for x, giving your answers to one decimal place.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences