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GATE CS 2003 Q4 — Combinatorics: Counting, Recurrence Relations, Generating Functions

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Let AA be a sequence of 88 distinct integers sorted in ascending order. How many distinct pairs of sequences, BB and CC are there such that

  1. each is sorted in ascending order,
  2. BB has 55 and CC has 33 elements, and
  3. the result of merging BB and CC gives AA
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Answer: C

Let the sorted sequence AA of 88 distinct elements be represented as A={a1,a2,…,a8}A = \{a_1, a_2, \dots, a_8\} where a1<a2<⋯<a8a_1 < a_2 < \dots < a_8.

We need to form two sequences, BB of size 55 and CC of size 33, such that:

  1. Both BB and CC are sorted in ascending order.
  2. Merging BB and CC yields the original sequence AA.

Since the merge of BB and CC results in AA, the elements of BB and CC must partition the elements of AA. That is, B∪C=AB \cup C = A and B∩C=∅B \cap C = \emptyset.

Because both BB and CC must be sorted in ascending order, the order of elements in each sequence is uniquely determined once the elements themselves are chosen. For example, if we choose any 55 elements from AA to form BB, there is exactly one way to arrange them in ascending order. The remaining 33 elements will form CC, and they also have exactly one valid ascending arrangement.

Thus, the number of distinct pairs of sequences (B,C)(B, C) is simply the number of ways to choose 55 elements out of 88 for sequence BB:

(85)=8!5!×3!=8×7×63×2×1=56\binom{8}{5} = \frac{8!}{5! \times 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

Therefore, there are 5656 distinct pairs of sequences.

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