Question

• A. The quantity on the left is greater
• B. The quantity on the right is greater
• C. Both are equal
• D. The relationship cannot be determined without further information

Hint:

Remember the property \(C(n, r) = C(n, n-r)\). It can simplify calculations. For instance, calculating \(C(8, 5)\) is easier as \(C(8, 3)\) because it involves fewer terms in the numerator.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves selection without regard to order. When Sunita or Shelly buys a set of pens, the order in which they pick them doesn't matter. This is a combination problem. We need to calculate the number of combinations for each case and compare them.
Step 2: Key Formula or Approach:
The number of ways to choose \(r\) items from a set of \(n\) distinct items is given by the combination formula: \[ C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} \] An important property of combinations is \(C(n, r) = C(n, n-r)\).
Step 3: Detailed Explanation:
For Column A:
Sunita has to buy 5 pens from 8 available companies. This is a problem of selecting 5 companies out of 8. Here, \(n = 8\) and \(r = 5\). Number of ways = \( C(8, 5) \). Using the property \(C(n, r) = C(n, n-r)\), we have \(C(8, 5) = C(8, 8-5) = C(8, 3)\). \[ C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56 \] So, Sunita can make her purchase in 56 ways.
For Column B:
Shelly has to buy 4 pens from 7 available companies. This is a problem of selecting 4 companies out of 7. Here, \(n = 7\) and \(r = 4\). Number of ways = \( C(7, 4) \). Using the property \(C(n, r) = C(n, n-r)\), we have \(C(7, 4) = C(7, 7-4) = C(7, 3)\). \[ C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35 \] So, Shelly can make her purchase in 35 ways.
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 56
Quantity B = 35
Therefore, Quantity A is greater than Quantity B.