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23 Oct 2020, 00:00
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Difficulty:
95%
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Question Stats:
25% (02:48) correct
75%
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wrong
based on 75
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A bag contains 6 chips numbered 1-6 respectively. If these chips are distributed among 4 students such that each student gets at-least one chip, how many different ways the chips can be distributed?
a. 480
b. 1080
c. 1200
d. 1560
e. 1800
a. 480
b. 1080
c. 1200
d. 1560
e. 1800
23 Oct 2020, 22:10
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I chose the method below as this is what came to my mind first. There may be a shorter approach.
Case 1: One person (Say A) takes 3 chips and each of the others take 1 chip each
Number of ways that A chooses 3 out of 6 chips = \(^6C_3 = 20\)
Number of ways that B chooses 1 out of the remaining 3 chips = \(^3C_1 = 3\)
Number of ways that C chooses 1 out of the remaining 2 chips = \(^2C_1 = 2\)
D gets the last chip in 1 way.
The number of ways to do the above = 20 * 3 * 2 = 120
There are 4 ways in which A or B or C or D could get 3 chips.
Therefore Total ways for case 1 = 120 * 4 = 480
Case 2: Two persons (Say A and B) take 2 chips each and each of the others take 1 chip each
Number of ways that A chooses 2 out of 6 chips = \(^6C_2 = 15\)
Number of ways that B chooses 2 out of the remaining 4 chips = \(^4C_2 = 6\)
Number of ways that C chooses 1 out of the remaining 2 chips = \(^2C_1 = 2\)
D gets the last chip in 1 way.
The number of ways to do the above = 15 * 6 * 2 = 180
The number of combinations of 2 people getting 2 chips each and the others getting 1 each = \(^4C_2 = 6\)
(AB, AC, AD, BC, BD or CD)
Therefore Total ways for case 2 = 180 * 6 = 1080
Total Possible ways = 480 + 1080 = 1560
Option D
Arun Kumar
Case 1: One person (Say A) takes 3 chips and each of the others take 1 chip each
Number of ways that A chooses 3 out of 6 chips = \(^6C_3 = 20\)
Number of ways that B chooses 1 out of the remaining 3 chips = \(^3C_1 = 3\)
Number of ways that C chooses 1 out of the remaining 2 chips = \(^2C_1 = 2\)
D gets the last chip in 1 way.
The number of ways to do the above = 20 * 3 * 2 = 120
There are 4 ways in which A or B or C or D could get 3 chips.
Therefore Total ways for case 1 = 120 * 4 = 480
Case 2: Two persons (Say A and B) take 2 chips each and each of the others take 1 chip each
Number of ways that A chooses 2 out of 6 chips = \(^6C_2 = 15\)
Number of ways that B chooses 2 out of the remaining 4 chips = \(^4C_2 = 6\)
Number of ways that C chooses 1 out of the remaining 2 chips = \(^2C_1 = 2\)
D gets the last chip in 1 way.
The number of ways to do the above = 15 * 6 * 2 = 180
The number of combinations of 2 people getting 2 chips each and the others getting 1 each = \(^4C_2 = 6\)
(AB, AC, AD, BC, BD or CD)
Therefore Total ways for case 2 = 180 * 6 = 1080
Total Possible ways = 480 + 1080 = 1560
Option D
Arun Kumar
General Discussion
23 Oct 2020, 22:45
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QuantMadeEasy
Apart from the way shown above, a quick method would be..
Let us distribute chips without any restrictions = \(4^6\)....Each chip can go to any of the 4 students.
Now this total has ways in which the chips have been distributed between just 3 of them or just 2 of them or just 1.
We have to subtract these ways..
To cater for above colored portion, let us SUBTRACT -
1) Amongst 3 of them..
Without restriction - 4C3*3^6
But these ways subtract the ways it is distributed between 2 and 1 more than ONCE.
example - Let us look at just A.
So A will be part of ABC, ABD and ACD, making him part of group of 2s twice.
ABC will give AB and AC
ABD will give AB and AD
ACD will give AC and AD
That is AB, AC, AB, AD, AC and AD
Thus the group of 2s is getting subtracted TWICE. Also, A will be considered single thrice, once in each of ABC, ABD and ACD. So it is getting subtracted thrice.
To cater for above colored portion, let us ADD
2) Between 2 of them
So let us add the way it will be distributed between two to cater for the EXTRA subtraction. => \(4C2*2^6\)
But again A will be part of AB, AC and AD, so A will get added THREE times, leading to NO subtraction at all.
To cater for above colored portion, let us SUBTRACT
3) Just 1
\(4C1*1^6\)
Answer --
\(4^6-4C3*3^6+4C2*2^6-4C1*1^6\)
\(= 4096-4*729+6*64-4*1\)
\(=4096-2916+384-4\)
\(=1560\)
D
