here the question asks only for selection and not for an arrangement of the selected nos. This is where I got confused. pick 3 or 5 elements, remaining elements will remain sorted since all are distinct nos.

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## 4 Answers

Best answer

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: $C$

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

Correct Answer: $C$

### 1 comment

Thanks @ Debashish Deka

here the question asks only for selection and not for an arrangement of the selected nos. This is where I got confused. pick 3 or 5 elements, remaining elements will remain sorted since all are distinct nos.

here the question asks only for selection and not for an arrangement of the selected nos. This is where I got confused. pick 3 or 5 elements, remaining elements will remain sorted since all are distinct nos.

### 7 Comments

You're counting same thing twice na. If you've picked up 3 elements for C, then by default 5 left out elements would be for set B.

Let's say we've three elements instead of 8 = {1,2,3}

For B you chose two elements out of 3, 3C2= {1,2} or {1,3} or {2,3} (not 3P2 since you can't rearrange them) So for C, you've got only one choice = {3}, {2}, {1} ) So total possible combinations are 3.

Say you chose elements for C first i.e., {1}, {2}, {3}

For B again, you've got to choose {2,3}, {1,3}, {1,2} only.

So you can see, if you consider choosing B and C as separate events, you'd have counted same possibilities twice.

Whenever such confusions occur, just try breaking the problem into smaller one.

Let's say we've three elements instead of 8 = {1,2,3}

For B you chose two elements out of 3, 3C2= {1,2} or {1,3} or {2,3} (not 3P2 since you can't rearrange them) So for C, you've got only one choice = {3}, {2}, {1} ) So total possible combinations are 3.

Say you chose elements for C first i.e., {1}, {2}, {3}

For B again, you've got to choose {2,3}, {1,3}, {1,2} only.

So you can see, if you consider choosing B and C as separate events, you'd have counted same possibilities twice.

Whenever such confusions occur, just try breaking the problem into smaller one.

In this question it is important to note that combination sequences do not have order.

for B sequences possible would be ->n-1C4+n-2C4+n-3C4+n-4C4 if you cal you get ->56

for C sequences possible would be ->n-1C2+n-2C2+n-3C2+n-4C2+n-5C2+n-6C2→ if you cal you get ->56

everytime you pick from the remainning elements ,you get another order so consider that order to be in increasing order .

now small observation here is for all B sequences there should be set C such that on merging you get A here you can think that for every possible sequence of B of 5 elements there will be a sequence of 3 elements in C such that on B intersection C is empty or null ,hence these sequence will also be 56 ,hence the answer would be 56 only

for B sequences possible would be ->n-1C4+n-2C4+n-3C4+n-4C4 if you cal you get ->56

for C sequences possible would be ->n-1C2+n-2C2+n-3C2+n-4C2+n-5C2+n-6C2→ if you cal you get ->56

everytime you pick from the remainning elements ,you get another order so consider that order to be in increasing order .

now small observation here is for all B sequences there should be set C such that on merging you get A here you can think that for every possible sequence of B of 5 elements there will be a sequence of 3 elements in C such that on B intersection C is empty or null ,hence these sequence will also be 56 ,hence the answer would be 56 only

→ 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**