TCS Company Numerical Ability Permutation and Combination

In how many ways can 7 different object be divided among three persons so that either one or two of them do not get any object?

Read Solution (Total 4)

Case 1: Two guys don't get anything. All items go to 1 person. Hence, 3 ways.

Case 2: 1 guy doesn't get anything. Division of 7 items among 2 guys. Lets say they are called A & B.

a) A gets 6. B gets 1. 7 ways.
b) A gets 5. B gets 2. 21 ways.
c) A gets 4. B gets 3. 35 ways.
d) A gets 3. B gets 4. 35 ways.
e) A gets 2. B gets 5. 21 ways.
f) A gets 1. B gets 6. 7 ways.
Total = 126 ways to divide 7 items between A & B.
Total number of ways for case 2 = 126*3 = 378 (Either A, or B, or C don't get anything).

Answer = Case 1 + Case 2 = 3 + 378 = 381
8 years agoHelpfull: Yes(18) No(2)

Ans : 381

A B C
7 0 0 3 ways
4 3 0 35*6 = 210 ways
6 1 0 7*6 = 42 ways
5 2 0 21*6 = 126 ways

Total = 3+210+42+126 = 381 ways
8 years agoHelpfull: Yes(7) No(0)
No of ways distributing objects when 1 person does not get any object is = 3*126 i.e 378 nd
When all the 7 objects will be given to 1 person ..it can be selected in three ways .So 378+ 3 = 381 (ans)
8 years agoHelpfull: Yes(2) No(0)
7c3-(7c1+7c2)= 35-(7+21)=7. Plz check if it would be right or not.
8 years agoHelpfull: Yes(1) No(7)

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