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3 votes
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The number of ways of selecting at least one Indian and at least one American for a debate from a group comprising $3$ Indians and $4$ Americans and no one else?
in Combinatory recategorized by
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3 Answers

6 votes
6 votes

We can use inclusion-exclusion principle (IEP). Well, IEP was not part of this test, but still.
 

#(At least one Indian and at least one American) = #(Total ways) – #(no Indian or no American)

#(Total ways) = 2^7. (We have 2 choices for each person –> either choose or reject) = 128

#(No Indian or no American) = #(no Indian) + #(no American) – #(no Indian and no American)

                                             = 2^4 (choose/reject from Americans only) + 2^3 (choose/reject from Indians only) – 1 (choose no one)

                                             = 16 + 8 -1 = 23

 

Hence, #(at least one Indian and at least one American) = 128 – 23 = 105.

5 votes
5 votes

NOTE that we want "at least one Indian and at least one American for a debate".. So, from $3$ Indians, we can choose any non-empty subset, hence, $2^{\wedge} 3-1=7$ ways.

Similarly, from $4$ Americans, we can choose any non-empty subset, hence, $2^{\wedge} 4-1=15$ ways.
Total we have $7\times 15=105$ ways

$\text{Method 2(Lengthy):}$

There are $3$ Indians in the group. We have to select at least $1$ Indian. i.e., we could select $1$ or $2$ or all $3$ Indians.

Similarly, there are $4$ Americans in the group. We have to select at least $1$ American. i.e., we could select $1$ or $2$ or $3$ or all $4$ Americans.

  • Number of ways of selecting $1$ Indian $=3$ choose $1=3 \mathrm{C} 1=3$ ways.
  • Number of ways of selecting $2$ Indians $=3$ choose $2=3 \mathrm{C} 2=3$ ways.
  • Number of ways of selecting all $3$ Indians $=3$ choose $3=3 \mathrm{C} 3=1$ way.
  • Hence, the number of ways of selecting one or more Indians $=(3+3+1)$ $=7$ ways.
  • Number of ways of selecting $1$ American $=4$ choose $1=4 \mathrm{C} 1=4$ ways.
  • Number of ways of selecting $2$ Americans $=4$ choose $2=4 \mathrm{C} 2=4 ! 2 ! * 2 != 6$ ways.
  • Number of ways of selecting $3$ Americans $=4$ choose $3=4 \mathrm{C} 3=4 \mathrm{C} 1=4$ ways.
  • Number of ways of selecting all $4$ Americans $=4$ choose $4=4 \mathrm{C} 4=1$ way Hence, the number of ways of selecting one or more Americans $=(4+6$ $+4+1)=15$ ways.

So, the number of ways of selecting at least one Indian and at least one American $=7 \times 15=105$ ways.

edited by
0 votes
0 votes

INDIAN | AMERICAN

1 | 1

1 | 2

1 | 3

1 | 4

….similarly for 2,3 Indians : 

(3C1 + 3c2 + 3C3) * (4c1 + 4C2 + 4c3 + 4C4) simple.

Answer:

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