Latest TCS Aptitude Question SOLUTION: In how many ways 7 different objects can be divided among 3 persons so that either one or two of them do not get any object.

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Permutation and Combination

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40038997) TCS Numerical Ability question Keep an EYE

In how many ways 7 different objects can be divided among 3 persons so that either one or two of them do not get any object.

Asked In TCS charu (9 years ago)

Unsolved Read Solution (8)

Is this Puzzle helpful?   (12)   (1) Submit Your Solution Permutation and Combination

Read Solution (8) :

RAKESH  (9 years ago)

if 2 of them don't get any, all object goes to 3rd person => 3 ways(A/B/C)

if 1 of them don't get any, 7 objects goes to rest 2 persons (AB,BC,CA)
this can be in 3 ways
(1,6)=> 7c1
(2,5)=> 7c2
(3,4)=> 7c3
(4,3)=> 7c4
(5,2)=> 7c5
(6,1)=> 7c6

(7c1+7c2+7c3+7c4+7c5+7c6)= 2^7-2 = 126
total ways = 3*126 = 378

Total = 3 + 378 = 381

Is this solution Helpfull? Yes (46) | No (3)  

Gokul  (9 years ago)

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

Is this solution Helpfull? Yes (12) | No (1)  

shreyansh  (9 years ago)

IT CAN BE ALSO DONE IN THS WAY
first case - 7,0,0 / 0,7,0 / 0,0,7 these are three ways
second case -we have to keep only one zero i.e. to divide 7 objects to 2 of them which is 7p2 and we have to choose 2 out of three which is 3c2 so answer is
3*7p2*3c2=381

Is this solution Helpfull? Yes (4) | No   

MUNISH YADAV  (9 years ago)

The possible cases in which either one or two of them do not get any object is:
x$$,$$x,$x$,xx$,$xx,x$x i.e 6 different cases
x means no object received
$ means at least one object received
Now for xx$,$xx,x$x cases where only 1 person receives the object, there is only 1 way of dividing the objects i.e. we have to give all the 7 objects to 1 person only. So, for these three cases we can divide the objects in 3 ways.
Now for x$$,$$x,$x$ cases where only 2 persons receive the objects, there can be different possibilities to divide the 7 objects amongst them. It can be done in these manner:
(1,6)(6,1)i.e 6*1+1*6 = 12 ways
(2,5)(5,2)i.e 2*5+5*2 = 10 ways
(3,4)(4,3)i.e 3*4+4*3 = 24 ways
So, total such ways = 12+10+24 = 46
Thus, 3+46 = 49 ways (Ans)

Is this solution Helpfull? Yes (3) | No (5)  

SUMIT  (9 years ago)

abe Rakesh tere ko koi or suject ata hai ki bs maths me hi tees maar khan hai......

Is this solution Helpfull? Yes (2) | No (3)  

Shagun  (9 years ago)

Rakesh.. I don't understand how u write (1,6) => 7c1 ??

Is this solution Helpfull? Yes (1) | No (1)  

MUNISH YADAV  (9 years ago)

@GOKUL thanks bhai

Is this solution Helpfull? Yes | No   

sanyogita holkar  (9 years ago)

28 for the condition wen 1 of them didnt get is 7, for the condition 2 of them didnt get is 21

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.