In the row, first and last place shall be taken by girls
No. of ways of filling first place and last place out of 4 girls = 4x3 = 12
No. of ways of filling remaining places = 5x4x3x2x1 = 120 (because 2 girls are already placed in 2 positions remaining 5 persons are remaining)
Total no. of ways = 12 x 120 =1440
Answer (D) 1440

The number of ways we can choose 2 girls from 4 to take the first and last positions is 4C2 = (4 x 3)/2! = 12/2 = 6. Once the two girls, say, A and B, are picked, there are two ways to switch them, since A can be the first position and B the last or B can be the first position and A the last. The 5 people in the middle can be arranged in 5! = 120 ways. Thus, the total number of ways we can arrange these 7 people so that the first and last positions are taken by girls is:

6 x 2 x 120 = 1,440

Answer: D
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=>

There are four ways (girls) to fill the first position and three ways (girls left over) to fill the last position. The remaining five positions can be filled by the five remaining children (three boys and two girls) in 5!=120 ways.

By the multiplication rule, there are 4 * 3 * 120 = 1,440 different lines possible.

Therefore, the answer is D.

Answer: D
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