The previous two posts
discussed a key idea behind counting number of options for the question of type a + b + c = r. Now, let us see a general version of this.
For a1 + a2 + a3 ….ar = n. The number of solutions where a1,a2,a3….ar all take natural numbers is (n-1) C (r-1).For a1 + a2 + a3 ….ar = n. The number of solutions where a1,a2,a3….ar can take all whole number values is (n+r-1) C (r-1).
Similar questions on the same theme
The proofs are just an extension of the idea discussed in the two posts. Now, this kind of question can get asked in different frameworks. Few examples are given below.
1. 10 identical toys need to be placed in 3 distinct boxes. In how many ways can this be done?
2. Mark needs to pick up exactly 40 fruits for his family. He has to pick at least 8 Apples, at least 6 Oranges and at least 5 mangoes. He should not pick up any other fruit. In how many ways can this be done?
- October 29, 2012 Questions and solutions on Permutation and Combination
Permutation and Combination.
1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?
2. In how many ways 11 identical toys be […] Posted in Combinatorics
- November 5, 2012 Permutations and Combinations Question
Sum of three whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist? (This is a variant on the previous question that can be found here). Will post the […] Posted in Combinatorics
- November 15, 2010 CAT Number Theory – Modulus Had sent an entry on Euler's phi function earlier , and thought it would be best to follow this up with a slightly more detailed post on modulus arithmetic.
If we divide a number N by […] Posted in Number Theory
- November 15, 2010 Fermat’s Little Theorem and Euler’s Phi function
Fermat's Little Theorem states that if p is prime then a^p - a is a multiple of p. In other words,
a ^ (p-1) = 1 mod p, whenever a is not a multiple of p
The proof for this […] Posted in Number Theory