Six discrete mathematics student:Bahirah,Dinesh,Johan,Adam,Felina and Syahirah are going to stand in a line. How many ways can they stand if:

a)Syahirah must be in the third position?
b)Dinesh must be second and Fellina third?
c)Boys and girls alternate, with boy starting the line?
d)The first three positions are boysand the last three are girls?
e)Adam can't be on eitherend of the line?

Number of arrangements of n persons in n positions is n!.

Example, 3 people, 3 places, there are 3!=6 possible arrangements.

a) If Syahirah mush be in the third position, there are 5 people left for 5 places. Use the formula above.

b)If Dinesh must be second, and Felina third, there are only 4 people left for 4 positions. Use the formula above.

c) If boys and girls alternate, with boy starting the line (assuming the line is directed, i.e. there is head and tail of the line), then boys are in the 1st, 3rd and 5th position, and girls in the remaining positions.
There are 3 places for 3 boys, and 3 places for 3 girls.
Multiply the number of arrangements for boys, and that of girls to get the answers. Multiplication because the arrangement of boys is (assumed to be) independent of that of girls.

d) Same idea as c) above, figure it out.

e) If Adam is at the beginning of the line, there are 5 positions left for 5 people. Same if he is at the end.
Calculate the number of arrangements for 6 people in six positions, then subtract the number of arrangements that Adam is at the head and tail of the line.

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To find the number of ways the students can stand in a line given the conditions, we can use the principle of permutations.

a) Syahirah must be in the third position.
We can fix Syahirah in the third position, leaving 5 students to arrange in the remaining 5 positions.
Number of ways = 5! = 120 ways

b) Dinesh must be second and Fellina third.
We can fix Dinesh in the second position and Fellina in the third position. The remaining 4 students can be arranged in the remaining 4 positions.
Number of ways = 4! = 24 ways

c) Boys and girls alternate, with a boy starting the line.
Since there are 3 boys and 3 girls, we can think of them as two separate groups and arrange them.
Number of ways to arrange boys = 3! = 6 ways
Number of ways to arrange girls = 3! = 6 ways
Total number of ways = Number of ways to arrange boys * Number of ways to arrange girls = 6 * 6 = 36 ways

d) The first three positions are boys, and the last three are girls.
We can fix the positions for boys in the first three positions and girls in the last three positions. The remaining student can be placed in the remaining position.
Number of ways to arrange boys = 3! = 6 ways
Number of ways to arrange girls = 3! = 6 ways
Number of ways to arrange the remaining student = 1 way
Total number of ways = Number of ways to arrange boys * Number of ways to arrange girls * Number of ways to arrange the remaining student = 6 * 6 * 1 = 36 ways

e) Adam can't be on either end of the line.
Adam cannot occupy the first and last positions. So, we can fix any position for Adam (other than the first and last) and arrange the remaining students.
Number of ways = 4! = 24 ways

Therefore,
a) There are 120 ways they can stand.
b) There are 24 ways they can stand.
c) There are 36 ways they can stand.
d) There are 36 ways they can stand.
e) There are 24 ways they can stand.