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Math, 25.02.2021 09:15 hajuyanadoy
Illustrate the following and determine the number of permutations. 1. Arranging 4 pots with different plants in a row
2. Forming a four-digit ATM pin.
3. Securing a motorcycle with a three-digit combination lock using the numbers 1, 2, 3 and 6
4. Displaying 3 identical small vases, 1 figurine, and a photo frame in a row
5. 3 girls sitting around a circular table​

Sagot :

abcdefdfgfghh

Answer:

1: 16

2: 4940

3: 24

4: 25

5: 9

Step-by-step explanation:

A permutation is a method of calculating the number of possible outcomes. It follows the following general formula;

[tex]_nP_r=\frac{n!}{(n-r)!}[/tex]

There (n) is the number of objects, and (r) is the number of objects selected.

1. Arranging 4 pots with different plants in a row

In order to solve this, one needs two pieces of information, the number of objects, and the number of objects selected. One is given the number of objects; (4), but when the problem states "in a row" it never specifies how many plants are in a row. Thus, let one assume that a "row" can have an infinite amount of space, but in this case, only (4) space will be used. Therefore there are (4) objects with (4) objects selected. However, the drawback is that the combination formula doesn't work when the two parameters (n) and (r) are the same. Hence, to solve this special case, one simply multiplies the two numbers to get the answers:

[tex]n*r\\\\=4*4\\\\=16[/tex]

2. Forming a four-digit ATM pin

One is given that there are (4) digits in the ATM pin, this is the number of objects selected. One is also given that number of objects, there are (10) digits including (0). Set up the permutation and solve;

[tex]_nP_r=\frac{n!}{(n-r)!}[/tex]

[tex]_1_0P_4=\frac{10!}{(10-4)!}\\\\=\frac{10!}{6!}\\\\=\frac{10*9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1}\\\\=10*9*8*7\\\\=4940[/tex]

3. Securing a motorcycle with a three-digit combination lock using the numbers (1), (2), (3), and (6).

There are (4) digits to choose from on the lock. But there are (3) numbers that can be selected.

[tex]_4P_3=\frac{4!}{(4-3)!}\\\\=\frac{4!}{1!}\\\\=\frac{4*3*2*1}{1}\\\\=4*3*2*1\\\\=24[/tex]

4. Displaying 3 identical small vases, 1 figure, and a photo frame in a row.

There are (5) objects, and (5) spaces (read problem (1) for an explanation for the objects being put in a row). Thus, this is a special case; multiply the two numbers to get the result;

[tex]n*r\\=5*5\\=25[/tex]

5. 3 girls sitting around a circular table

There are (3) subjects, and (3) spaces in this problem. Apply the same logic applies to a row in this problem. Therefore, this is another special case; multiply the two numbers to get the result;

[tex]n*r\\=3*3\\=9[/tex]

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