Find the value of the combination. 5C2

The Correct Answer and Explanation is :

The value of the combination 5C2\mathbf{^5C_2} is: 5C2=5!2!(5−2)!=5!2!⋅3!=1202⋅6=12012=10^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10

Explanation

In mathematics, a combination is a way of selecting items from a larger set where order does not matter. This is different from a permutation, in which the order does matter.

The notation nCr^nC_r, read as “n choose r”, represents the number of ways to choose r items from a group of n distinct items, without considering the order of selection. The formula for calculating combinations is: nCr=n!r!(n−r)!^nC_r = \frac{n!}{r!(n – r)!}

Here, “!” (factorial) denotes the product of all positive integers up to that number. For example, 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120.

For the problem 5C2^5C_2, we’re choosing 2 items from a set of 5. Applying the formula: 5C2=5!2!(5−2)!=1202!⋅3!=1202⋅6=12012=10^5C_2 = \frac{5!}{2!(5-2)!} = \frac{120}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10

So, there are 10 unique ways to choose 2 items from 5 items when the order doesn’t matter.

To understand this intuitively, suppose we have 5 items labeled A, B, C, D, and E. The possible unique combinations of 2 items are:

There are 10 such pairs, confirming our calculation.

This concept is widely used in probability, statistics, and real-world scenarios such as forming committees, lottery combinations, and selecting teams, where the arrangement of selected items is not important—only the group composition matters.

Explanation

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