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44 votes
44 votes

Let $A$ be a sequence of $8$ distinct integers sorted in ascending order. How many distinct pairs of sequences, $B$ and $C$ are there such that

  1. each is sorted in ascending order,
  2. $B$ has $5$ and $C$ has $3$ elements, and
  3. the result of merging $B$ and $C$ gives $A$
  1. $2$
  2. $30$
  3. $56$
  4. $256$
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4 Answers

Best answer
55 votes
55 votes
If you pick any $3$ numbers in the given order from the array(sorted) remaining $5$ elements are already sorted. You can not change the relative position of those $5$ elements because they are distinct and already sorted. So no of ways $= {}^8C_3.$

Same argument holds for picking up $5$ elements initially. No of ways $= {}^8C_5.$

Correct Answer: $C$
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34 votes
34 votes

answer - C

select any $3$ elements from given $8$ elements - $^8C_3$

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19 votes
19 votes
The question actually is based on combinations not merging.Firstl y two arrays are taken : B having 3 elements and C having 5 elements sorted in ascending order and then merged into a single array A having 8 elements.We have to find no of distinct sequences. 8 elements can be arranged in 8! ways out of that 3 elements of array B can be arranged in 3! ways and 5 elements of C can be arranged in 5! ways. Now, No. of distinct sequences = 8!/3! 5! = 56 Ans.
0 votes
0 votes

A can have sequence of 8 distinct integers which are sorted in ascending order.
→ If we are pick 3 elements from 8 sequence integers then remaining 5 elements are already in ascending order. After merging these elements then it gives A.
→ No. of possibilities of choosing 3 elements from total of 8 = $^8C$$_3$
= 8!/3!5!
= 8 * 7
= 56

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Answer:

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