Combinations Practice Quiz: Ace Your Exam

The combination formula is n!/(k!(n-k)!), which counts the number of ways to choose k items from n without considering the order of selection. It divides the total arrangements by the number of ways to order the chosen k items.

5 choose 2 is computed as 5!/(2!3!) which equals 10. This calculation involves dividing 120 by (2*6), yielding the correct number of combinations.

The number of ways to choose 2 items from 4 is calculated by C(4,2)=4!/(2!*2!) which equals 6. This process uses the combination formula where order does not matter.

The binomial coefficient calculates the number of ways to choose k items from n, without considering the order. It is a key concept in combinations and is given by the formula n!/(k!(n-k)!).

In combinations, the order in which items are selected is irrelevant. This distinguishes them from permutations, where the order is significant.

C(7,3) is computed using the formula 7!/(3!*4!), which equals 35. This represents the number of ways to choose 3 items from 7 without regarding the order.

The number of 3-element subsets from 9 elements is given by the combination 9C3, which calculates to 84. This formula counts the selections without considering the order.

Using the combination formula, C(12,4)=12!/(4!*8!) equals 495. This number reflects all possible committees that can be formed where the order of selection is not important.

By the symmetry property of combinations, C(8,3) is equal to C(8,5), and both compute to 56. This property follows because choosing 3 items to include is equivalent to choosing 5 items to exclude.

Select 3 boys out of 5 and 2 girls out of 4, then multiply the results: 5C3 * 4C2 = 10 * 6 = 60. This exemplifies the multiplication principle in combinatorics.

Since the positions are identical, the selection order does not matter, and the number of ways is computed by 10C2, which equals 45. This avoids the overcounting that would occur if order were considered.

Choosing 1 item from 10 can be done in exactly 10 ways, confirming that C(10,1)=10. This simple case reinforces a basic property of combinations.

A key property of combinations is that C(n, k) equals C(n, n-k). This symmetry occurs because choosing k items to include is equivalent to choosing n-k items to leave out.

Each row in Pascal's Triangle sums to a power of 2. For the 5th row (starting at row 0), the sum is 2^5, which equals 32.

The number of ways to choose 2 items from 6 is given by the combination 6C2, which equals 15. This is a direct application of the basic combination formula.

Since one member is required, the problem reduces to choosing the remaining 3 members from 11, calculated as C(11,3)=165. This adjustment is a common technique when restrictions are applied in combinations.

This problem counts the number of surjective functions from 7 books to 3 students. Using inclusion-exclusion, the total is calculated as 3^7 - 3*(2^7) + 3*(1^7) = 1806.

The number of 5-card hands is given by the combination 52C5, which equals 2,598,960. This approach disregards the order of cards, focusing solely on the selection.

First, arrange the 5 unrestricted individuals in 5! ways, then choose 3 gaps from the 6 available to insert the specified individuals, and finally arrange these 3 in 3! ways. This yields 5! * C(6,3) * 3! = 14400, ensuring the 3 do not sit adjacent.

Using the stars and bars method, the number of solutions is given by C(7+3-1, 3-1) = C(9,2) which equals 36. This method accounts for the distribution of 7 identical items among 3 variables.