What is the total number of ways in which Dishu can distribute 9 disti

Question

What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift?

A. 2592*8!
B. 144*8!
C. 72*8!
D. 36*8!
E. 72*7!

TestPrepUnlimited · Accepted Answer

Label each girlfriend A to H.

Set the scenario for A getting two gifts, then A has 9C2 possible choices and the rest of the seven girlfriends each get one gift out of 7 so that is 7! ways to distribute the rest. Then for scenario A we have  9C2 *7!  distributions.

We need to repeat this scenario for all 8 girlfriends so we have  8*9C2*7! = 8*9*8/2 *7! = 36*8! . There I think the choices from this question is flawed.

The post from above fails to consider the cases where A gets gift 1 first then gift 2 last, and another case where A gets gift 2 first then gift 1 last. These two should count as the same case so we need to divide by 2 with the method in said post.

MarmikUpadhyay · Answer

Number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift =?

First of all, let us distribute 8 distinct gifts to 8 distinct girlfriends.

Selection of 8 gifts from 9 distinct gifts can be done in  9 C 8  = 9 ways.
Distribution of those 8 gifts to 8 girlfriends can be done in 8! ways.

So 8 distinct gifts can be distributed to 8 different girlfriends in 9 * 8! ways such that all of them receives exactly one gift.

The remaining one gift can be given in 8 different ways, as there are 8 girlfriends.

Total number of ways to distribute 9 distinct gifts among 8 distinct girlfriends = 9 * 8! * 8 = 72 * 8! ways

So, correct answer is option  C .

NirajMore117 · Answer

Let us consider two cases
Case1: Each gf gets a single gift
Therefore by fcp it can be done in 9! ways

Case 2:One gf gets a double gift
By fcp distribution of 8 gift can be done in 8! ways. The remaining one gift can be distributed in 8 ways. 
Thus total ways in this case is 8*8!

Total number of ways=9!+8*8! =9*8!+8*8! 
=(9*8)8! =72*8!

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sumitkrocks · Answer

What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift?

Lucky Dishu can do this by

choosing one GF out of 8 who will get two gifts and choosing which two gifts that girlfriend will get( rest 7 will get leftover 7 gifts i.e each will get one)

Choose one GF *Choose two Gifts* distribute 7 remaining to 7 Gfs

8C1*9C2*7! =36*8!

No choice matches the outcome

ScottTargetTestPrep · Answer

Solution:

Since there are 9 distinct gifts and 8 girlfriends and each of the girlfriends gets at least one gift, it must be true that 1 girlfriend gets 2 gifts and all the other girlfriends get 1 gift each.

To solve this problem, we can use smaller numbers, for example, we can say, there are 4 distinct gifts and 3 girlfriends since this doesn’t change the fact that 1 girlfriend will get 2 gifts while all the other girlfriends get 1 gift each.

We can let A, B, C and D be the gifts, we can divide the 4 gifts into 3 groups in the following ways:

A-B-CD, A-C-BD, A-D-BC, B-C-AD, B-D-AC and C-D-AB

For each of the ways above and let’s take A-B-CD, we can distribute them to the 3 girlfriends in the following ways:

A-B-CD, A-CD-B, B-A-CD, B-CD-A, CD-A-B, CD-BA

Therefore, there are a total of 6 x 6 = 36 ways. Now, let’s analyze each factor despite both being 6. The first factor 6 is 4C2 since basically it is the number of ways 2 gifts can be chosen from 4 to give to the girlfriend who will receive the 2 gifts. The second factor 6 is 3! since basically it is the number of ways one can distribute the gifts amongst the 3 girlfriends once a partition of gifts is formed. In other words, if there are 4 gifts and 3 girlfriends, the number of ways the gifts can be distributed is 4C2 x 3!. Therefore, if there are 9 gifts and 8 girlfriends, the number of ways the gifts can be distributed is 9C2 x 8! = 36 x 8!.

Answer: 36 x 8!

Regor60 · Answer

The only way the given answer makes sense is if the meaning of "distribute" refers to how Dishu hands out the items rather than their resulting distribution.

So Dishu can distribute to the lucky personer who gets two items item A first and then item B second or item B first and item A second.

This accounts for two ways while the resulting distribution remains the same.

Let's demonstrate with a simple example:

3 gifts A,B,C and 2 people, 1,2

The approach used by a number of respondents that matches the provided answer is:

Select 2 gifts from the 3 to give to each of the 2 people:

3!/2 = 3

These gifts can be distributed 2! ways

So 2 distinct gifts can be distributed to 2 distinct people:

3*2 = 6 ways

The remaining gift can be given 2 ways to the 2 different people

So the total ways is:

2*6 = 12 ways

Now, lets work through the actual distribution scenarios:

1 A
2 BC

1 B
2 AC

1 C
2 AB

1 AB
2 C

1 AC
2 B
 
1 BC
2 A

So, a total of 6 ways, half the number suggested by the process yielding the provided answer.

The approach that matches provided answer suggests that:

1 BC
2 A

is distinct from

1 CB
2 A

Does it matter to the recipients of two gifts what order they receive them ?

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Kinshook · Answer

Asked: What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift?

Number of ways to distribute 8 distinct gifts one each among his 8 distinct girlfriends = 9C8 *8! = 9*8! = 9!
Number of ways to distribute last distinct gift among his 8 distinct girlfriends = 8C1 = 8 
Total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift = 8*9*8! = 72*8!
 
IMO C

Legallyblond · Answer

why didnt we use inclusion - exclusion formula here? can you please explain??????

Bunuel · Answer

We don’t use inclusion-exclusion here because it’s unnecessary and overcomplicates the problem. If we wanted to use inclusion-exclusion, the count would be:

Total functions (no restriction) = 8^9
 
Subtract cases where at least one gets nothing:
 
= 8^9 - C(8,1)*7^9 + C(8,2)*6^9 - C(8,3)*5^9 + C(8,4)*4^9 - C(8,5)*3^9 + C(8,6)*2^9 - C(8,7)*1^9

This alternating sum continues until we include all possible restrictions, making it long and messy. The final result would simplify to 36 * 8!, but it’s much easier to get that directly with basic counting logic:

Choose the girlfriend who gets 2 gifts: 8 ways 
 Choose which 2 gifts she gets: C(9,2) = 36 
 Distribute the remaining 7 gifts to the 7 others: 7! ways

Total = 8 * 36 * 7! = 288 * 7! = 36 * 8!.

Legallyblond · Answer

the answer is 72*8! right?

Bunuel · Answer

The answer is 36 * 8!.

Legallyblond · Answer

rst of all, let us distribute 8 distinct gifts to 8 distinct girlfriends.

Selection of 8 gifts from 9 distinct gifts can be done in 9C8 = 9 ways. 
 Distribution of those 8 gifts to 8 girlfriends can be done in 8! ways.

So 8 distinct gifts can be distributed to 8 different girlfriends in 9 * 8! ways such that all of them receives exactly one gift.

The remaining one gift can be given in 8 different ways, as there are 8 girlfriends.

Total number of ways to distribute 9 distinct gifts among 8 distinct girlfriends = 9 * 8! * 8 = 72 * 8! ways  
 If you do this question from this approach you get 72*8!.... why is this method wrong? and what would you suggest as in how should we think in such a problem henceforth? should i think from the point of view of 2 gifts?

Bunuel · Answer

Theat method double-counts.

In 9 * 8! * 8, each case where one girlfriend gets 2 gifts is counted twice (for example, gifts A and B to the same person are treated separately as AB and BA). That’s why the correct count divides by 2, giving 36 * 8!.

Legallyblond · Answer

i think we can start from first which girl get 2 gifts in 8c1 ways and which 2 gifts--- 8c1*9c2*7!

Adit_ · Answer

Is this the hardest Quants question in terms of success rate? Only 2%??
I tried solving using options initially and realised later that there was something wrong with my approach and tried doing using usual methodical approach:

9 distinct things distributed to 8 distinct people.

First 9 things can be distributed by 2,1,1,1,1,1,1,1 to 8 distinct girlfriends.

The above setup is fixed as no other way can we have atleast 1 gift to everybody involved.

Now there are a few combinations here:

First choose 9c7 items for all the 1s in the setup.
= 36.

Now we can arrange the above setup in 8C7 ways = 8 ways.

Since all are distinct we can also shuffle those gifts around those 1s anyway among the 1s. The remaining 2 gifts will always remain with the ones having those 2 gifts.
So its enough to arrange the 7 gifts among 1s = 7! ways.

Now we simply need to multiply all of them together:

36*8*7! ways = 36*8! ways.

Answer:  Option D

What is the total number of ways in which Dishu can distribute 9 disti

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What is the total number of ways in which Dishu can distribute 9 disti

Prep Toolkit

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