1. There are 10 athletes joining the Baguio Marathon Event. How many ways can the first, second, and third placers be chosen?
There are 10 options for the first placer, 9 options for the second placer (since one athlete has already been chosen as first placer), and 8 options for the third placer (since two athletes have already been chosen). Therefore, the total number of ways to choose the first, second, and third placers is:
$10 \times 9 \times 8 = \boxed{720}$ ways.
2. Four boys and three girls are to arrange themselves to form a line.
(a) The boys stand together, and the girls stand together?
We can treat the group of boys as a single entity and the group of girls as another single entity. Then, we can arrange these two entities in the line in 2! = 2 ways (either the boys are in front of the girls or the girls are in front of the boys). Within each group, the boys can be arranged in 4! = 24 ways and the girls can be arranged in 3! = 6 ways. Therefore, the total number of arrangements is:
We can think of the line as consisting of a "slot" for a boy, a "slot" for a girl, a "slot" for a boy, and so on. There are 4! = 24 ways to arrange the boys in their slots and 3! = 6 ways to arrange the girls in their slots. Therefore, the total number of arrangements is:
$4! \times 3! = \boxed{144}$ arrangements.
3. There are 10 participants to a barangay meeting to talk about safety measures to be implemented during the pandemic. In how many ways can these 10 barangay participants arrange themselves in a row if three of them should stay together?
We can treat the group of three participants as a single entity. There are 8! ways to arrange the remaining 8 entities, and 3! ways to arrange the three entities within their group. Therefore, the total number of arrangements is:
$8! \times 3! = \boxed{20,!160}$ arrangements.
4. A coat hanger has four knobs, and each knob is to be painted by a distinct color. If six different colors of paint are available, how many ways can the knobs be painted?
There are 6 options for the color of the first knob, 5 options for the color of the second knob (since one color has already been used), 4 options for the color of the third knob, and 3 options for the color of the fourth knob. Therefore, the total number of ways to paint the knobs is:
5. A husband and wife with their three children are to pose for a family picture.
(a) They can take any position?
There are 5 entities (husband, wife, and 3 children) that can be arranged in a row in 5! = 120 ways. Therefore, the total number of arrangements is:
$5! = \boxed{120}$ arrangements.
(b) The mother and the father stay at both ends?
The mother and father can be arranged in 2 ways (either the mother is on the left and the father is on the right or vice versa). Within each of these arrangements, the 3 children can be arranged in 3! = 6 ways. Therefore, the total number of arrangements is:
$2 \times 3! = \boxed{12}$ arrangements.
(c) In how many ways can they be arranged in two rows if the parents will stay at the front row?
The parents can be arranged in 2 ways (either the mother is in front and the father is behind her, or vice versa). For each of these arrangements, there are 3! ways to arrange the children in the front row. There are then 2 options for which child is in front in the second row, and 2! ways to arrange the other two children in the second row. Therefore, the total number of arrangements is:
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1. There are 10 athletes joining the Baguio Marathon Event. How many ways can the first, second, and third placers be chosen?
There are 10 options for the first placer, 9 options for the second placer (since one athlete has already been chosen as first placer), and 8 options for the third placer (since two athletes have already been chosen). Therefore, the total number of ways to choose the first, second, and third placers is:
$10 \times 9 \times 8 = \boxed{720}$ ways.
2. Four boys and three girls are to arrange themselves to form a line.
(a) The boys stand together, and the girls stand together?
We can treat the group of boys as a single entity and the group of girls as another single entity. Then, we can arrange these two entities in the line in 2! = 2 ways (either the boys are in front of the girls or the girls are in front of the boys). Within each group, the boys can be arranged in 4! = 24 ways and the girls can be arranged in 3! = 6 ways. Therefore, the total number of arrangements is:
$2 \times 4! \times 3! = \boxed{288}$ arrangements.
(b) The boys and the girls stand alternately?
We can think of the line as consisting of a "slot" for a boy, a "slot" for a girl, a "slot" for a boy, and so on. There are 4! = 24 ways to arrange the boys in their slots and 3! = 6 ways to arrange the girls in their slots. Therefore, the total number of arrangements is:
$4! \times 3! = \boxed{144}$ arrangements.
3. There are 10 participants to a barangay meeting to talk about safety measures to be implemented during the pandemic. In how many ways can these 10 barangay participants arrange themselves in a row if three of them should stay together?
We can treat the group of three participants as a single entity. There are 8! ways to arrange the remaining 8 entities, and 3! ways to arrange the three entities within their group. Therefore, the total number of arrangements is:
$8! \times 3! = \boxed{20,!160}$ arrangements.
4. A coat hanger has four knobs, and each knob is to be painted by a distinct color. If six different colors of paint are available, how many ways can the knobs be painted?
There are 6 options for the color of the first knob, 5 options for the color of the second knob (since one color has already been used), 4 options for the color of the third knob, and 3 options for the color of the fourth knob. Therefore, the total number of ways to paint the knobs is:
$6 \times 5 \times 4 \times 3 = \boxed{360}$ ways.
5. A husband and wife with their three children are to pose for a family picture.
(a) They can take any position?
There are 5 entities (husband, wife, and 3 children) that can be arranged in a row in 5! = 120 ways. Therefore, the total number of arrangements is:
$5! = \boxed{120}$ arrangements.
(b) The mother and the father stay at both ends?
The mother and father can be arranged in 2 ways (either the mother is on the left and the father is on the right or vice versa). Within each of these arrangements, the 3 children can be arranged in 3! = 6 ways. Therefore, the total number of arrangements is:
$2 \times 3! = \boxed{12}$ arrangements.
(c) In how many ways can they be arranged in two rows if the parents will stay at the front row?
The parents can be arranged in 2 ways (either the mother is in front and the father is behind her, or vice versa). For each of these arrangements, there are 3! ways to arrange the children in the front row. There are then 2 options for which child is in front in the second row, and 2! ways to arrange the other two children in the second row. Therefore, the total number of arrangements is:
$2 \times 3! \times 2 \times 2! = \boxed{48}$ arrangements.