· 10 objects, 5 people. Each person gets exactly 2 objects.

· Each person values every 10 objects. The object that a person receives is known as a bundle. He will have his own total value, and can have total value of others’ received objects.

· This comparison of values of person’s own value with others received bundle ‘value for him’, determines ENVY.

If A values own bundle=10, and values B’s bundle=12, (values B’s bundle more than his own), then, A envies B.

Given:

· 3 people value own bundle = 16. No one values own bundle more than 16.

· D values own bundle at odd number. ⟹ D ≠ 16. All other bundles are even.

· Point 6: Person who values own bundle < 16, envies person who values own bundle = 16.

No one envies others, means that People who value own bundle=16, do not envy other people’s bundles who value their own < 16.

· We can also mark value of object=10, as received by that person. Also, The object received by one person can be marked red, as it cannot be used by other persons.

	o1	o2	o3	o4	o5	o6	o7	o8	o9	o10	Total
Amar	4	9	9	3	7	3	8	7	9	5		even
Barat	5	9	7	5	5	3	6	8	10	8	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Barat.
Charles	8	8	8	3	6	4	5	8	9	6		even
Disha	8	8	8	5	5	3	6	4	9	8		odd
Elise	6	8	9	5	6	5	6	3	7	10	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Elise.

· Since o7 has been received by Barat, then, it cannot be received by Elise.

· Using point 6 as discussed above: A’s value for B’s bundle is 8+9=17, which is greater than the maximum possible total for any person. Therefore, A must have his bundle value < 16, so that A envies B, and condition of point 6 is satisfied.

· Therefore, A < 16, B=16, E=16, D is odd, then the third person having total=16 must be Charles.

	o1	o2	o3	o4	o5	o6	o7	o8	o9	o10	Total
Amar	4	9	9	3	7	3	8	7	9	5	<16	even
Barat	5	9	7	5	5	3	6	8	10	8	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Barat.
Charles	8	8	8	3	6	4	5	8	9	6	16	even
Disha	8	8	8	5	5	3	6	4	9	8		odd
Elise	6	8	9	5	6	5	6	3	7	10	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Elise.

· Charles needs total=16. From among the numbers available for Charles, we can make 16 using only 8s. There are 8 in o1, o2, o3, and o8. But we have 2 conditions in point 2 and point 3.

o o1, o2, o3 are given to 3 different people.

o o1 and o8 are given to different people.

Therefore, to get 16, we must select o8 and one among o2 or o3.

· For Amar, to make 16 from 2 object values, we need either 2 even values, or 2 odd values. There is only 1 even object value available. Therefore, o1 is not received by Amar.

	o1	o2	o3	o4	o5	o6	o7	o8	o9	o10	Total
Amar	4	9	9	3	7	3	8	7	9	5	<16	even
Barat	5	9	7	5	5	3	6	8	10	8	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Barat.
Charles	8	8	8	3	6	4	5	8	9	6	16	even
Disha	8	8	8	5	5	3	6	4	9	8		odd
Elise	6	8	9	5	6	5	6	3	7	10	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Elise.

· If C has received o2. B will value C’s bundle (o2 + o8) with value=17. This means B will envy C, which is not possible due to point (6).

Therefore, C must have received o3 and o8 only.

· For D, to get odd total, he needs 1 even and 1 odd value. Possible even value are of: o1, o2. Possible odd values are of: o4, o4, o6.

E needs o1 or o5 to make total value=16.

We can make 2 cases for o1 received either by D or E. (A cannot receive o1, because it will need another even value to make total value even)

Case 1: o1 is received by E. Then o2 will be received by D.

	o1	o2	o3	o4	o5	o6	o7	o8	o9	o10	Total
Amar	4	9	9	3	7	3	8	7	9	5	<16	even
Barat	5	9	7	5	5	3	6	8	10	8	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Barat.
Charles	8	8	8	3	6	4	5	8	9	6	16	even
Disha	8	8	8	5	5	3	6	4	9	8		odd
Elise	6	8	9	5	6	5	6	3	7	10	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Elise.

In this case, A values D’s o2 at value=9. A’s own value of bundle (3+3 or 3+7) will be less than A valuing D’s bundle. Therefore, this case will be eliminated.

Case 2: o1 is received by D. E will receive o5 to make total value 16. o2 will be received by A.

	o1	o2	o3	o4	o5	o6	o7	o8	o9	o10	Total
Amar	4	9	9	3	7	3	8	7	9	5	<16	even
Barat	5	9	7	5	5	3	6	8	10	8	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Barat.
Charles	8	8	8	3	6	4	5	8	9	6	16	even
Disha	8	8	8	5	5	3	6	4	9	8		odd
Elise	6	8	9	5	6	5	6	3	7	10	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Elise.

Now, A has to get one of o4 or o6, D will get the other.

· If A gets o4, D will get o6. D’s bundle value in this case = 8+3 = 11. D will value A’s bundle = 8+5 = 13. This means D will envy A, which is not possible according to point (6).

Therefore, A will get o6, and D will receive o4.

Final table is as follows:

	o1	o2	o3	o4	o5	o6	o7	o8	o9	o10	Total
Amar	4	9	9	3	7	3	8	7	9	5	12	even
Barat	5	9	7	5	5	3	6	8	10	8	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Barat.
Charles	8	8	8	3	6	4	5	8	9	6	16	even
Disha	8	8	8	5	5	3	6	4	9	8	13	odd
Elise	6	8	9	5	6	5	6	3	7	10	16	even, with 1 value=10, to get even total, less than 16, all values except 6 cannot be used for Elise.

Object o4 was given to Disha. Therefore, answer is option(a): Disha.

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