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Official PYQ · 2025

GATE CS 2025 Set 1 Q10 — General Aptitude

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A shop has 44 distinct flavors of ice-cream. One can purchase any number of scoops of any flavor. The order in which the scoops are purchased is inconsequential. If one wants to purchase 33 scoops of ice-cream, in how many ways can one make that purchase?

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Answer: B

This is a classic problem of combinations with repetition (also known as the stars and bars problem or finding the number of non-negative integer solutions).

We need to choose r=3r = 3 scoops of ice-cream from n=4n = 4 distinct flavors. Since we can choose multiple scoops of the same flavor and the order of purchase does not matter, we can model this as:

x1+x2+x3+x4=3x_1 + x_2 + x_3 + x_4 = 3

where xi≥0x_i \ge 0 represents the number of scoops chosen of flavor ii.

The number of non-negative integer solutions is given by the formula:

(n+r−1r)=(4+3−13)=(63)\binom{n + r - 1}{r} = \binom{4 + 3 - 1}{3} = \binom{6}{3}

Calculating the value:

(63)=6×5×43×2×1=20\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

Alternative Method (Case-by-Case Analysis):

Let the four flavors be A,B,C,A, B, C, and DD. We can categorize the selections as follows:

  1. All 3 scoops of the same flavor (e.g., AAAAAA):
    There are (41)=4\binom{4}{1} = 4 ways.
  2. 2 scoops of one flavor and 1 of another (e.g., AABAAB):
    Select the flavor for 2 scoops in (41)=4\binom{4}{1} = 4 ways, and the flavor for 1 scoop in (31)=3\binom{3}{1} = 3 ways.
    Total ways = 4×3=124 \times 3 = 12 ways.
  3. All 3 scoops of different flavors (e.g., ABCABC):
    Select 3 distinct flavors out of 4 in (43)=4\binom{4}{3} = 4 ways.

Adding the ways from all cases:

Total ways=4+12+4=20\text{Total ways} = 4 + 12 + 4 = 20

Thus, there are 20 ways to make the purchase.

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