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In how many ways can 3 non-negative integers be chosen such that a + b + c = 10?

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  • Last Updated : 22 Dec, 2021

Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

Permutation Formula

In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

nPr = (n!)/(n – r)!

Here,

n = group size, the total number of things in the group 

r = subset size, the number of things to be selected from the group

Combination

A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Combination Formula

In combination r things are picked from a set of n things and where the order of picking does not matter.

nCr = n!⁄((n-r)! r!

Here,

n = Number of items in set

r = Number of things picked from the group

In how many ways can 3 non-negative integers be chosen such that a + b + c = 10?

Solution:

x1 + x2 + x3 + x4 + ………. + xr = n

then, no. of ways with non-negative integer value = n+r-1Cr-1

a + b+ c = 10

n = 10, r = 3

No. of ways with non-negative integer value = 10+3-1C3-1

                                                                      = 12C2 = 12!/2!10!

                                                                      = 66 ways

​Similar Questions

Question 1: If a + b + c + d = 20, how many unique, non-negative integer solutions exist for (a, b, c, d)?

Solution:

x1 + x2 + x3 + x4 + ………. + xr = n

then, no. of ways with non-negative integer value= n+r-1Cr-1

a + b+ c + d = 20

n = 20, r = 4

No. of ways with non-negative integer value = 20+4-1C4-1

                                                                      = 23C3 = 23!!/3!20!

                                                                      = 1771 ways

Question 2: If a + b + c + d = 20, how many positive integer solutions exist for (a, b, c, d)?

Answer:

We assign at least a value of 1 to a, b, c, d.

So, we can say a = a + 1, b = b+ 1, c = c + 1, d = d + 1 where a, b c, d are positive integers

= a+ 1 + b+ 1 + c + 1 + d + 1 = 20

= a + b + c + d = 20 – 4

= a + b + c + d = 16

Number of solutions = (16+(4-1)) C(4-1) 

                                            = 19C3  = 19!/3!16!

                                 = 969

Question 3: If a + b + c + d=15, how many unique, non-negative integer solutions exist for (a, b, c, d) such that a > 5 and b > 2.

Solution:

We allocate at least a value of 5 to a and 2 to b

So, a = a + 5 and b = b + 2

= a + 5 + b + 2 + c + d = 15

= a+ b + c + d = 8

Number of solutions = (8+4-1)C(4-1) 

                                 = 11C3 = 11!/3!8!

                                 = 165

Question 4: If a + b + c + d = 20, how many unique, non-negative integer solutions exist for (a, b, c, d) such that a > b.

Solution:

Let us first understand the state where a = b

If a = b = 0, c + d = 20. This has 21 possibilities

If a = b = 1, c + d = 18. This has 19 possibilities

If a = b = 2, c + d = 16. This has 17 possibilities

.

.

If a = b = 10, c + d = 0. This has 1 possibility

So, the total number of a possibilities when a = b is 21 + 19 + 17 … + 1 = 11/2*(21 + 1) = 121

We know that the numeral of chances when a, b, c, and d are non-negative number is 1771. Out of these 1771 occurrence, in 121 occurrence a = b.

So, in 1771 – 121 = 1650 occurrence a is not equal to b.

In half of the above occurrence a will be greater than b whereas in the other half of the occurrence a will be less than b.

So, number of solutions where a > b is 1650/2 = 825

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