Serge is handing out snacks at the day care center. There are 15 children .He has seven apples, six oranges and two boxes of raisins. In how many different ways could the snack be distributed?
There are different ways to approach this problem, but one possible method is to use the multiplication principle of counting. This principle states that if a task can be broken down into a sequence of independent decisions, where each decision has a certain number of options, then the total number of ways to complete the task is the product of the number of options at each decision point.
In this case, we can break down the task of distributing snacks into three decision points: choosing which children get an apple, which children get an orange, and which children get a box of raisins. Since each child can only get one snack, and each snack can only be given once, these decisions are independent. Therefore, we can multiply the number of options at each decision point to get the total number of ways to distribute snacks:
Number of ways to choose which children get an apple = 15 options, since any of the 15 children can get an apple
Number of ways to choose which children get an orange = 14 options, since one child has already gotten an apple, so there are only 14 left to choose from
Number of ways to choose which children get a box of raisins = 13*12/2 = 78 options, since there are two boxes of raisins (which can be given to different or the same children), and we need to choose two children from the remaining 13 who have not yet gotten an apple or an orange (and order doesn't matter, so we divide by 2)
Therefore, the total number of ways to distribute snacks is:
15 * 14 * 78 = 16,380
So there are 16,380 different ways that Serge can distribute the snacks among 15 children using 7 apples, 6 oranges, and 2 boxes of raisins.
There are five different colours available in a babric that is suitable for a flag. How many different flags of three stripes, all vertical or all horizontal, (tricolours) can be made, assuming that no colour is repeated in a flag?
If there are five different colors available, we can choose the first stripe of the flag in 5 ways, since any of the colors can be used. For the second stripe, we can choose any of the four remaining colors. Finally, for the third stripe, we can choose any of the three colors remaining. Therefore, there are:
5 * 4 * 3 = 60
ways to make a tricolour flag with different colors for each stripe, assuming that the stripes are all vertical or all horizontal.
Note that if we allow diagonal stripes, or repeated colors, or different numbers of stripes, the number of possible flags would be different.
An Austrian village consists of five streets running north and southand three streets running east and west. A tourist sets out from the north-e ast corner of the village to reach the south-west corner. In how many different ways could the tourist wander these streets if she is always facing south or west?
If the tourist is always facing south or west, then at each intersection, she can only turn either left or right. Therefore, the problem can be reduced to counting the number of ways to arrange five L's (for left turns) and three R's (for right turns) in a row. The number of such arrangements is given by:
8!/(5!3!) = 56
This is because we have 8 total symbols (5 L's and 3 R's), and we want to count the number of distinct arrangements of these symbols, where the order matters but the repetitions of the same symbol do not. This can be calculated using the formula for combinations with repeated elements.
Therefore, there are 56 different ways for the tourist to wander these streets if she is always facing south or west.
Seven year old Sarah his nine crayons; three blue, one red, two green and three yellow.(a) In how many ways can she line up the crayons on her desk?(b)She is to pick up some of the crayons. How many different choices of some crayons could she make?
(a) If Sarah wants to line up all nine crayons on her desk, she can do so in 9!/(3!1!2!3!) = 60,480 ways. This is because there are 9 total spots for the crayons, but some of the spots are indistinguishable due to multiple crayons of the same color. Specifically, there are three indistinguishable spots for the blue crayons, one indistinguishable spot for the red crayon, two indistinguishable spots for the green crayons, and three indistinguishable spots for the yellow crayons. Therefore, the total number of distinct arrangements is the number of ways to permute the nine spots, divided by the number of indistinguishable arrangements of the same color crayons.
(b) To count the number of choices of some crayons, we can use the fact that for each crayon, Sarah can either include it in her selection or not. Therefore, the total number of choices is given by 2^9 = 512, since there are 9 crayons and she has 2 choices for each. However, this includes the case where she picks no crayons at all, which is not allowed. Therefore, the actual number of different choices of some crayons that she can make is 511.
Of the 21 young girls auditioning for the role of Annie, nine were good dancers who could not sign well enough, but five could sing and dance well. Assuming that all girls possessed at least one of these skills, how many girls must have been good singers without dancing skill?
Let's use the principle of inclusion-exclusion to solve this problem. We know that:
- 9 girls were good dancers but not good singers.
- 5 girls were good singers and dancers.
- There were 21 girls in total.
We want to find the number of girls who were good singers but not good dancers. Let S be the set of all girls who were good singers, and D be the set of all girls who were good dancers. Then we can use the formula:
|S \ D| = |S| - |S ∩ D|
where |X| denotes the size (number of elements) of set X. In other words, we subtract the number of girls who were both good singers and good dancers from the number of girls who were just good singers.
Now, |S| = 5, since there were 5 girls who were good at both singing and dancing. To find |S ∩ D|, we can use the formula:
|S ∩ D| = |S| + |D| - |S U D|
where |X U Y| denotes the size of the union of sets X and Y. In other words, we add the number of girls who were good singers and the number of girls who were good dancers, but we remove the number of girls who were good at both (to avoid double-counting).
We know |S| = 5 and |D| = 9, but what about |S U D|? This is just the size of the set of all girls, which is 21. Therefore, we have:
|S ∩ D| = 5 + 9 - 21 = -7
Wait, that doesn't make sense! The size of a set cannot be negative. The reason for this is that we have counted some girls twice: those who were both good singers and good dancers. Therefore, we need to subtract their number one more time, which gives:
|S ∩ D| = 5 + 9 - 2*5 = 4
Now we can use the formula for |S \ D|:
|S \ D| = |S| - |S ∩ D| = 5 - 4 = 1
So there was only one girl who was a good singer but not a good dancer.
To find the number of different ways the snacks can be distributed, we need to consider the number of choices for each snack and the total number of children.
Let's break it down step by step:
1. Start with apples:
- We have 7 apples, and they can be distributed among the 15 children. Each child can have 0, 1, or more apples.
- We can use a technique called "stars and bars" to find the number of ways to distribute the apples. In this case, there are 15 positions (children) and 7 dividers (apples).
- So, the number of ways to distribute apples is given by the formula: (15+7-1) choose (7) = 21 choose 7.
2. Move on to oranges:
- We have 6 oranges, and they can also be distributed among the 15 children.
- Similarly, using the stars and bars technique, the number of ways to distribute oranges would be: (15+6-1) choose (6) = 20 choose 6.
3. Finally, consider the raisins:
- We have 2 boxes of raisins, which means they can be distributed as a whole unit (not separating them).
- So, there are only two options for distributing the raisins - either one box goes to one child, and the other box goes to another child, or both boxes go to the same child.
4. To find the total number of ways to distribute all three snacks, we multiply the results from steps 1, 2, and 3:
- Total ways = [(15+7-1) choose (7)] * [(15+6-1) choose (6)] * 2
Now, let's calculate the result:
Total ways = (21 choose 7) * (20 choose 6) * 2
Using combinatorial notation, this simplifies to:
Total ways = C(21, 7) * C(20, 6) * 2
Calculating this expression will give you the final answer for the number of different ways the snacks can be distributed among the 15 children.