IAS 2020 — CSAT Maths

Correct Option

There are 5 distinct denominations available: ₹50, ₹100, ₹200, ₹500, and ₹2000.

The problem requires forming different sums by selecting at least three denominations at a time. This implies considering combinations of 3, 4, or 5 denominations from the given 5.

The number of ways to choose 'r' items from 'n' distinct items is given by the combination formula:

$$\left( {}^{n}C_{r} = \frac{n!}{r!(n - r)!} \right)$$(nCr​=r!(n−r)!n!​)

• Number of ways to choose exactly 3 denominations: $5C_{3} = \frac{5!}{3!(5 - 3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$5C3​=3!(5−3)!5!​=3!2!5!​=2×15×4​=10.
• Number of ways to choose exactly 4 denominations: $^{5}C_{4} = \frac{5!}{4!(5 - 4)!} = \frac{5!}{4!1!} = 5$5C4​=4!(5−4)!5!​=4!1!5!​=5.
• Number of ways to choose exactly 5 denominations: $^{5}C_{5} = \frac{5!}{5!(5 - 5)!} = 1$5C5​=5!(5−5)!5!​=1.

The total number of different sums is the sum of these possibilities:

Required different sums = $5C_{3} +^{5}C_{4} +^{5}C_{5}$5C3​+5C4​+5C5​ = 10 + 5 + 1 = 16.

Incorrect Options

• Option 2 (15): This value would be obtained if only combinations of 3 and 4 denominations were considered ($5C_{3} +^{5}C_{4} = 10 + 5 = 15$5C3​+5C4​=10+5=15), omitting the case where all 5 denominations are selected.

• Option 3 (14): This value does not correspond to a direct calculation based on the problem's criteria for selecting at least three denominations.

• Option 4 (10): This value represents only the number of ways to choose exactly 3 denominations ($5C_{3} = 10$5C3​=10), failing to account for combinations of 4 or 5 denominations as required by the "at least three" condition.