Question

for a set of three elements find the number of different subsets of 3 elements. use row 3 of pascals triangle to find the answer.
the number of different subsets of 3 elements for the set of three elements is .

Explanation:

Step1: Recall Pascal's triangle

Pascal's triangle has rows numbered starting from 0. The $n$-th row of Pascal's triangle gives the binomial coefficients $\binom{n}{k}$ for $k = 0,1,\cdots,n$. The formula for the elements in the $n$-th row of Pascal's triangle is $\binom{n}{k}=\frac{n!}{k!(n - k)!}$, where $n! = n\times(n - 1)\times\cdots\times1$.

Step2: Identify $n$ and $k$

We have a set of $n = 3$ elements and we want to find the number of subsets of $k=3$ elements. We look at the 3 -rd row of Pascal's triangle. The 3 -rd row of Pascal's triangle has the numbers 1, 3, 3, 1 which correspond to $\binom{3}{0},\binom{3}{1},\binom{3}{2},\binom{3}{3}$ respectively.

Step3: Calculate the binomial coefficient

We want $\binom{3}{3}$. Using the formula $\binom{3}{3}=\frac{3!}{3!(3 - 3)!}=\frac{3!}{3!0!}$. Since $0!=1$ and $3!=3\times2\times1 = 6$, we have $\binom{3}{3}=\frac{6}{6\times1}=1$.

Answer:

1