
If four boys and three girls are to be seated for a dinner such that no two girls and no two boys sit together. Find the number of ways in which they can be seated?
A.144
B.244
C.300
D.225
Answer
556.8k+ views
Hint: Here we will first write the possible situation of the seating arrangement of the boys and girls. Then we will use the concept of the permutation and solve the equation to get the number of ways in which they can be seated. The permutation is defined as the different ways in which a collection of items can be arranged and the order of arrangement matters.
Complete step-by-step answer:
Given that there are four boys and three girls are to be seated for a dinner such that no two girls and no two boys sit together.
Then there is the only possibility that on the first place boys are seated and on the second place a girl is seated and so on alternatively seated. In the end place i.e. seventh place a boy is seated.
Therefore the number of the ways is \[ = \left( {4!} \right) \times \left( {3!} \right)\]
Now we will simply expand the factorial terms in the above equation to get the value of the number of ways. Therefore, we get
Number of ways \[ = \left( {4 \times 3 \times 2 \times 1} \right) \times \left( {3 \times 2 \times 1} \right) = 144\]
Hence the number of ways in which they can be seated is 144.
So, option A is the correct option.
Note: Here we should note that for finding the number of ways for the particular case we use the permutation, not combination. As combinations may be defined as the various ways in which objects from a set may be selected and not arranged.
Factorial of a number is equal to the multiplication or product of all the positive integers smaller than or equal to the number. In addition, the factorial of 1 is always equal to 1, and the factorial of 0 is equal to 1. The factorial of a number is always positive, it can never be negative and the factorial of a negative number is not defined.
Complete step-by-step answer:
Given that there are four boys and three girls are to be seated for a dinner such that no two girls and no two boys sit together.
Then there is the only possibility that on the first place boys are seated and on the second place a girl is seated and so on alternatively seated. In the end place i.e. seventh place a boy is seated.
Therefore the number of the ways is \[ = \left( {4!} \right) \times \left( {3!} \right)\]
Now we will simply expand the factorial terms in the above equation to get the value of the number of ways. Therefore, we get
Number of ways \[ = \left( {4 \times 3 \times 2 \times 1} \right) \times \left( {3 \times 2 \times 1} \right) = 144\]
Hence the number of ways in which they can be seated is 144.
So, option A is the correct option.
Note: Here we should note that for finding the number of ways for the particular case we use the permutation, not combination. As combinations may be defined as the various ways in which objects from a set may be selected and not arranged.
Factorial of a number is equal to the multiplication or product of all the positive integers smaller than or equal to the number. In addition, the factorial of 1 is always equal to 1, and the factorial of 0 is equal to 1. The factorial of a number is always positive, it can never be negative and the factorial of a negative number is not defined.
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