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Numerical Ability
Permutation and Combination
In how many ways can 7 distinct objects be divided among three people so that either one or two of them do not get any object?
Read Solution (Total 7)
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- 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 - 9 years agoHelpfull: Yes(32) No(9)
- assume there are three persons A,B ,C
now consider following two cases
1.
A do not get anything
in this case 7 objects can be distributed between B and C as follows
B C
1 6 - total ways are 7!/1!*6!=7
2 5 -total ways are 7!/2!*5!=21
3 4 - total ways are 7!/3!*4!=35
4 3 - total ways are 7!/4!*3!=35
5 2 - total ways are 7!/5!*2!=21
6 1 - total ways are 7!/6!*1!=7
hence total ways of distributing objects between B and C = 126
similarly for (A c)and(AB) each =126
CASE2-
two persons do not get anything
in thi case there will be three waya these are
007 ,070 ,700
combining both the cases
total no of ways =126*3+3=381 - 8 years agoHelpfull: Yes(17) No(0)
- assuming person C is empty handed
1st object can be distributed to 2 person
similarly 2nd,3rd,4th,5th,6th,7th
no of ways =2*2*2*2*2*2*2
now assuming B &a as empty handed
total ways=3*2^7-3 =381
we subtracted 3 ,becoz there is conidtion when only 1 person get the object is repeated twice i.e 6 instead of 3 - 9 years agoHelpfull: Yes(3) No(1)
- 36 is the answer
- 9 years agoHelpfull: Yes(1) No(6)
- 7c2+7c1=28 is answr
- 9 years agoHelpfull: Yes(1) No(3)
- 1st situation:
when 2 of them don't get any object i.e. all 7 are given to 1 of them . so there can be 1 out of 3 which gets all so , 3 ways.
2 situation :
when one of them doesn't get any of the objects i.e. 7 objects are divided between two of them . Since these 7 are distinct objects so choices are also made that which object is given to which guy of the two.
so,
7c1+7c2+7c3+7c4+7c5+7c6=126 ways.
Now this is the no. of ways in which objects are divided between 2 of them
Now we can also select these two of the 3
so we have 3 choices
so final ans is:
3*126+3=381.
- 8 years agoHelpfull: Yes(1) No(1)
- 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 - 5 years agoHelpfull: Yes(0) No(0)
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