Quant Question Of The Day: 45
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Counting
There are five boys A, B, C, D, E who are to form a team to accomplish a particular task. How many distinct teams can be formed that contains at least one of the boys such that A & B don’t want to be in same team and also C & D don’t want to be in same team?
1. 15
2. 16
3. 17
4. 18
5. None of these
There are exactly 3 ways for A & B {(i) both A & B not selected, (ii) A selected & B not selected, (iii) A not selected & B selected} to be selected in the team.
Same ways there are exactly 3 ways for C & D to be selected in the team.
And there are exactly 2 ways for E i.e. to be selected or not selected in the team.
So total cases becomes = 3×3×2 = 18. But it includes the case when none of them is included in the team which we need to subtract to get the final answer as 18 – 1 = 17.
Solve our previous ‘questions of the day’
Quant Question Of The Day: 44
Quant Question Of The Day: 43
A,B,C,D,E,AC,BC,CE,DE,AD,BD,ACE,AE,BE,ADE,CBE,DBE = 17
SORRY, THE CORRECT ANSWER IS A,B,C,D,E,AC,BC,CE,DE,AD,BD,ACE,AE,BE,ADE,CBE,DBE= 17
17
16
One boy each – 5
Pair of boys – 8
Group of 3 boys – 4
Total distinct teams = 17
17
Correct me if m wrong about my method
A-3 ways
B-3
C-3
D-3
E-5 ways
Sum of all 17
7
17
Why 1 is subtracted in the end?