GATE CSE 2025 | Set 1 | GA | Question: 10

Question

A shop has $4$ 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 $3$ scoops of ice-cream, in how many ways can one make that purchase?

• A. $4$
• B. $20$
• C. $24$
• D. $48$

Comments

i respect your thoughts, but here i only put this link for learning combination with repetition concept, not to directly copy their question...
this is very interesting and intuitive playlist for discrete math, this prof. teaches very intuitive way... so that...

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To find boxes in a question always remember that, a box can take or hold 2 values at the same time in real scenario.

here a scoop can be any of the 4 flavours but in real it will just take one flavour

but here a flavour can be taken to scoop1 scoop2 at real scenario without any error, so flavours are boxes, scoops are balls

so distinct flavours so DB and scoops sequence is invalid so scoops are identical objects

so IODB. which is star bar problem, choose r slots from n+r-1 slots.

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suppose 4 colors are b,r,g,y  then
BRG,RGY,BRY,BGY     4                     -without repetition

BBB,BBY,BBR,BBG    4
RRR,RRY,RRG,RRB   4                    -   with repetition
YYY.......................       4
GGG......................      4
              TOTAL           4*5=20
 

Answer 1

This is similar to a combination with repetition: \[ x_1 + x_2 + x_3 + x_4 = 3 \] where: \[ x_1 \quad \text{denotes the number of scoops of the 1st flavor} \] \[ x_2 \quad \text{denotes the number of scoops of the 2nd flavor} \] \[ x_3 \quad \text{denotes the number of scoops of the 3rd flavor} \] \[ x_4 \quad \text{denotes the number of scoops of the 4th flavor} \] The combination formula for this is: \[ \binom{n + r - 1}{r} \] For this \( 6C3 \), it is written as: \[ \binom{6}{3} = 20 \]

Comments

4C1+4C2+4C3=20       1.for three ice cream have same flavour
                                      2.two are same and one has different flavour
                                     3.three has different flavour

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How can be the value of 4c1 + 4c2 + 4c3 be 20 ?? Brother @Rahmat_Ali.

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@Monalisa mam

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Exactly and this is method is called Star Bars Problem.
This is taught by Deepak Sir in detailed.
It is IODB Template

Answer 2

This is a Simple approch to this question...

A person has to select 3 scoops from 4 differnent flavours let ( A,B,C,D)

1. he can select 3 scoops of same flavour = 4 ways

2. he can select 2 scoops of same flavour and 1 scoop of different flavour

        suppose he selects  A A then for the third scoop he had 3 choices = 3

      in the same way for BB , CC and DD he has = 3+3+3 = 9 choices

total for A,B,C,D = 3+9 = 12 choices

3. he can choose 3 scoops of 3 different flavous.. which is 4C3..= 4

 so total no of ways he can select 3 scoops from 4 different flavours is = 4 + 12 + 4 = 20.

 so, the answer is B.

Answer 3

This Belongs to IODB Template , this IODB Templatee can be used to solve 5 answers , onee is Combination with Repetition.... so basically, 

"The order in which the scoops are purchased is inconsequential" ==> Combination 

"One can purchase any number of scoops of any flavor" ==> Repetition

 

So this can be solved using star-bar problem , 

We want=stars=identical = 3 scoops

box=category=kind=4 flavours

 

3 stars,4 boxes => 3 stars,3 bars => (3+3)! / (3!*3!) => 6C3 => 20

 

Answer 4

refer this

Comments

can you tell me why you had multiplied with 2 in 2 diferent scoops

 

Answer 5

Suppose we have, Vanilla, Strawberry, Butterscotch, Chocolate

A team of 3 flavours

But flavours can be used repeatedly (unlike a team of people)

V S B = B S V (will be same in this question)

→ 4C3 won’t work (takes an item only once)

→  4P3 won’t work too (as it counts orders)

We are looking for combinations where order doesn’t matter but allowing repetitions

_ _ _

Case 1: All same flavours

→  4 ways

Case 2: 2 same flavours

→  _ _ 2 spots can be filled in 4 ways and last spot can be filled in 3 ways

Total is 12 ways

Case 3: All different flavours

Equivalent to picking 3 flavours without replacement

→  4C3 = 4!/3!1! = 4 ways

Total ways = 4 + 12 + 4 = 20 ✅ Option B

Answer 6

lets say we have r scoops okay 
and lets say x1 , x2 , x3 ... xn are the no of scoops of each flavours . 
than can u tell me x1 + x2 + x3 ....+xn = ?? 
that must be equal to r right ?? b/c we have r scoops in total 
now i gues you have head of stars and bars problem in which we distributed some same number of r items in two n disticnt bags 
like say we have r stars *** and n bars ||  so yeah for seprating into n bags bars must be n-1 ?? like for seprating 4 starts into two bags we can have ****  and a | in between so if a bag have x items and other have y x+y must be ??? i guess now you know the answer it's r beacuse total items are r . now the total ways to do that would be equal to no of ways to arange n-1 bars in total r+n-1 items . r+n-1 c n-1 . 
likewise that we have 3 scoops okay and we have 4 flavours 
so total scoops are 3 which represent stars okay we need to divide that in 4 flavours ( bag) which represents the bars so total bars must be 4-1 = 3 
now total items 3 + 3 ( stars and bars ) and we need to select 3  
6c3 = 20 
btw we can also do it by one more method 
we can divide the three scoops as 
1. all of same flavour ( 4 chances )
2. two of same flavour (4c2*2)
3. all different (4 chances ) 
add those also gives us 20