Question

The number of subsets containing at most $5$ elements that can be picked from a set of $8$ elements is
A. $256$
B. $219$
C. $220$
D. None of these

Answer 1

Hint:Here we are going to use combination and permutations standard formula. $\begin{array}{l}
^nc{}_r = \frac{{n!}}{{r!(n - r)!}}\\
{}^np{}_r = \frac{{n!}}{{(n - r)!}}
\end{array}$

Complete step by step solution:
Given that: $5$ elements to be picked from $8$ elements-
If we pick any $5$ elements one at a time, then there would be $8$ different ways,$7$ different ways to pick the 2nd element, $6$ different ways to pick the 3rd element, $5$ different ways to pick 4th element and $3$ different ways to pick 5th element.
That, $\begin{array}{l}
{}^np{}_r = \frac{{n!}}{{(n - r)!}}\\
{}^np{}_r = \frac{{8!}}{{(8 - 5)!}}\\
{}^np{}_r = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}}\\
{}^np{}_r = 6720\\

\end{array}$
But order in which pick a particular subset doesn’t matter here, so we can divide this value by the number of ways of ordering $5$ items, that is $5!$ ,
We get, $\begin{array}{l}
{}^nc{}_r = \frac{{n!}}{{r!(n - r)!}}\\
{}^nc{}_r = \frac{{8!}}{{5!(8 - 5)!}}\\
{}^nc{}_r = \frac{{8!}}{{5!3!}}\\
{}^nc{}_r = \frac{{8 \times 7 \times 6 \times 5!}}{{5!3!}}\\
{}^nc{}_r = \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}\\
{}^nc{}_r = 56
\end{array}$ (Taking $5!$ From Numerator & denominator)

Hence, the given option D goes for the solution.

Additional information: Combinations is used if certain objects are to be arranged in such a way that the order of objects is not important whereas Permutation is an ordered combination- an act of arranging the objects or numbers in the specific order.


Note: Always use combination and permutations for these types of sums.