Permutation and Combination

Question

In how many ways can 3 non-negative integers be chosen such that a + b + c = 10 where a >= -1 , b >= -5 and c >= 3 ?

• A.  36
• B. 66
• C. 105
• D. None

Answer 1

You can first satisfy lower constraint then apply combination with unlimited repetition formula.

a+b+c=10+1+5-3=13

Now r=13 and n=3

Then answer is 105.

Comments

MY DOUBT :

In question it is already mentioned that a,b and c are 3 non-negative numbers . So Conditions a >= -1 , b >= -5  does not make sense. so ultimately question will be like this “ In how many ways can 3 non-negative integers be chosen such that a + b + c = 10 where c >= 3 ? “ 
now we can first satisfy lower constraint . Therefore answer will be 36 only.

Answer 2

for solving this problem you have to know about the star bar problem or (distributing  identical object in to non identical boxes)

in combinatorics .

In the question ,they are asking about non - negative  integer ,so we need to just care about the third condition ,i e, c>=3.

To satisfy this we give 3 unit value to c variable in first place ,so the equation becomes a + b + c= 7

Now there are no conditions , so we can directly apply Star bar equation then answer will be 

7+2C2 = 9C2

Comments

MY DOUBT :

In question it is already mentioned that a,b and c are 3 non-negative numbers . So Conditions a >= -1 , b >= -5  does not make sense. so ultimately question will be like this “ In how many ways can 3 non-negative integers be chosen such that a + b + c = 10 where c >= 3 ? “ 
now we can first satisfy lower constraint . Therefore answer will be 36 only.

Answer 3

To find the number of ways to choose three non-negative integers (a, b, and c) such that a + b + c = 10 with the given constraints (a >= -1, b >= -5, and c >= 3), we can use a combinatorial approach, specifically using stars and bars (balls and urns) method.

 

The stars and bars method allows us to distribute identical objects (stars) into distinct containers (bars). In this case, the stars represent the value of a, b, and c, and the bars represent the partitions that separate these values.

 

Let's denote the number of stars as 10, which corresponds to the sum of a, b, and c. We will introduce two additional variables s and t to consider the constraints:

 

1. s represents the offset for a: a' = a + s. To satisfy the constraint a >= -1, we set s = 1. So, a' = a + 1, and now a' >= 0.

2. t represents the offset for b: b' = b + t. To satisfy the constraint b >= -5, we set t = 5. So, b' = b + 5, and now b' >= 0.

 

Now, we have the equation a' + b' + c = 10, with a' >= 0, b' >= 0, and c >= 3. We can ignore the non-negativity constraints for a' and b' in this problem as they are guaranteed to be non-negative due to the constraints on a and b.

 

So, we have a simplified equation: a' + b' + c = 10 with a' >= 0, b' >= 0, and c >= 3.

 

Now, we can use stars and bars to find the number of ways to arrange these values.

 

Let's consider 10 stars (representing a' + b' + c = 10) and two bars (representing the partitions between a', b', and c):

 

Example: **|****|******| (corresponding to a' = 2, b' = 5, and c = 3)

 

Now, to satisfy c >= 3, we will distribute 3 stars to c before placing the bars:

 

Example: **|****|***|*****| (corresponding to a' = 2, b' = 5, and c = 5)

 

There are 7 objects (10 stars and 2 bars), and we need to choose 2 positions for the bars. This can be done in (7 choose 2) ways.

 

The number of ways to choose three non-negative integers a, b, and c, such that a + b + c = 10 and satisfying the given constraints, is:

 

Number of ways = (Number of ways to place the bars) = (7 choose 2) = C(7, 2) = 7! / (2! * (7 - 2)!) = 21.

 

Therefore, there are 21 ways to choose three non-negative integers a, b, and c satisfying the conditions a + b + c = 10, a >= -1, b >= -5, and c >= 3.

Answer 4

The given problem is a typical combination with unlimited repetition problem, with a slight added lower constraints on each of the three variables a, b and c. We can simply solve it in two steps : 

1. Filter out all the lower constraints by adding the given lower constraint values i.e. (-1) + (-5) + 3 = -3 and subtracting this value from the right hand side of the equation i.e 10 – (-3) = 13.   
2. Since it is a case of combination with unlimited repetition we can simply use the formula n-1+r C r. 
3. The final answer on solving is 3-1+13C13 => 15C13 => 15C2 => 105 → final answer.