Analyzing the Unrestricted Universe: S1
Imagine you have a bag of ten numbered balls, and you are asked to pick three. You can pick the same ball multiple times. This is the essence of S1={(i,j,k):i,j,k∈{1,2,…,10}}.
There are no restrictions on the selection. For the first position i, you have 10 choices; for the second position j, you have 10 choices; and for the third position k, you have 10 choices.
By the Fundamental Principle of Counting, the total number of triplets is:
Thus, Statement (A) is confirmed.
The Constrained Dance: S2
Now, consider S2={(i,j):1≤i<j+2≤10}. This is a dance of inequalities where we must respect the boundaries.
First, look at j+2≤10, which simplifies to j≤8. Since j must be at least 1, j can be any integer from 1 to 8.
For each j, the condition i<j+2 means i can be any integer from 1 up to j+1. This gives us (j+1) possible values for i.
We sum these values for j=1 to j=8:
n2=j=1∑8(j+1)=2+3+4+⋯+9
Using the arithmetic progression sum formula:
Statement (B) is solid.
The Elegance of Order: S3
Here is where many students stumble. S3={(i,j,k,l):1≤i<j<k<l≤10}. The strict inequality i<j<k<l is a gift.
It means that once you select any four distinct numbers from our set of ten, there is exactly one way to arrange them to satisfy the inequality. You do not need to worry about permutations; this is a pure selection problem.
We need to choose 4 numbers out of 10:
n3=10C4=4×3×2×110×9×8×7=210
Statement (C) claims 220, so it is false.
The Chaos of Permutations: S4
Finally, S4={(i,j,k,l):i,j,k,l are distinct elements in {1,2,…,10}}. Here, the order matters, and we are picking 4 distinct items.
This is a permutation problem. We have 10 choices for the first, 9 for the second, 8 for the third, and 7 for the fourth:
Checking Statement (D), we calculate:
It is true! We have successfully navigated the landscape of this problem. Remember, the key is to translate the notation into the right combinatorial tool.
