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Q: Composers choose instrument combinations when orchestrating a piece is this true or false?

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Since there are 3 numbers to choose from, there will be 2^3 = 8 combinations. This includes the null combinations which you may not wish to include in the count.

There are 36 possible combinations.

There are 220 combinations. Do you want to know why this is the answer? (It's worth knowing because this is a widely used principle in mathematics.) There are 10 digits to choose from, four of them are to be chosen at a time and a combination of four can include more any given digit more than once. So the total number of 4-digit combinations (without insisting that there be any 2s in them) is (10+4-1)choose 4= 13 choose 4 = 715. If 2s are _not_ allowed then there would be 9 digits to choose from and the total number of 4-digit combinations would be (9+4-1) choose 4 = 12 choose 4 = 495. Then the number of 4-digit combinations in which anywhere from 1 to 4 2s are allowed would be 715 - 495 = 220.

If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.

86,450

Related questions

True

yesThey are usually quite specific about which instruments (and how many of each) are used in their compositions.

he chose his instrument the trumpet because he wanted a brass instrument

I raise this qn

Since there are 3 numbers to choose from, there will be 2^3 = 8 combinations. This includes the null combinations which you may not wish to include in the count.

The answer is 4,960.

10

beacause it is my hobby

There are 36 possible combinations.

The number of 4 different book combinations you can choose from 6 books is;6C4 =6!/[4!(6-4)!] =15 combinations of 4 different books.

There are 220 combinations. Do you want to know why this is the answer? (It's worth knowing because this is a widely used principle in mathematics.) There are 10 digits to choose from, four of them are to be chosen at a time and a combination of four can include more any given digit more than once. So the total number of 4-digit combinations (without insisting that there be any 2s in them) is (10+4-1)choose 4= 13 choose 4 = 715. If 2s are _not_ allowed then there would be 9 digits to choose from and the total number of 4-digit combinations would be (9+4-1) choose 4 = 12 choose 4 = 495. Then the number of 4-digit combinations in which anywhere from 1 to 4 2s are allowed would be 715 - 495 = 220.

If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.