A is a set of three digit number whose sum of the digits is equal to 20 find A
Answers
992
answer...............
Hello guys Namaste,
A three-digit positive integer has the form 100h+10t+u, where the hundreds digit h satisfies the inequalities 1≤h≤9, the tens digit t satisfies the inequalities 0≤t≤9, and the units digit u satisfies the inequalities 0≤u≤9. Therefore, we wish to determine the number of solutions of the equation
h+t+u=12(1)
subject to these restrictions.
If we let h′=h−1, then 0≤h′≤8. Substituting h′+1 for h in equation 1 yields
h′+1+t+uh′+t+u=12=11(2)
Equation 2 is an equation in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of eleven ones. There are
(11+22)=(132)
such solutions since we must choose which two of the thirteen symbols (eleven ones and two addition signs) will be addition signs.
However, we have counted solutions in which h′>8, t>9, or u>9. We must exclude these.
Suppose h′>8. Then h′ is an integer satisfying h′≥9. Let h′′=h′−9. Then h′′≥0. Substituting h′′+9 for h′ in equation 2 yields
h′′+9+t+uh′′+t+u=11=2(3)
Equation 3 is an equation in the non-negative integers. Since a particular solution corresponds to the placement of two addition signs in a row of two ones, it has
(2+22)=(42)
solutions.
Suppose t>9. Then t is an integer satisfying t≥10. Let t′=t−10. Substituting t′+10 for t in equation 2 yields
h′+t′+10+uh′+t′+u=11=1(4)
Equation 4 is an equation in the non-negative integers with (32)=3 solutions, depending on which variable is equal to 1. By symmetry, there are also three solutions in which u′>9. No two of these restrictions cannot be violated simultaneously since 9+10=19>12. Thus, the number of three-digit positive integers with digit sum 12 is
(132)−(42)−2(32)=66