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pentagon [3]
3 years ago
8

Four different mathematics books, six different physics books, and two different chemistry books are to be arranged on a shelf.

How many different arrangements are possible if a) The books in each particular subject must all stand together b) Only the mathematics books must stand together
Mathematics
1 answer:
sveta [45]3 years ago
4 0

Answer:

a) 207,360 ways

b)8,709,120  ways

Step-by-step explanation:

Total number of subjects= 3

Number of mathematics books = 4

Number of physics books = 6

Number of chemistry books = 2

Part a:

Number of possible arrangement s= 3! 4! 6! 2!= 6*24*720*2= 207360 ways

In this we have included 3! because the books have to be placed in such a manner that the books in each particular subject must all stand together.

Part b:

Similarly the mathematics books must stand together so they are treated as

The total number of books will be

Physics books + Chemistry books+ Mathematics books= 6+2+1(treated as single) = 9

But there are 4 different mathematics books so these can be arranged in 4 ways.

Total number of ways in which these books can be arranged = 9! 4!= 8709120  ways.

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<em><u>Solution:</u></em>

<em><u>Given function is:</u></em>

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f(6) = 2(6) - 1 = 12 - 1 = 11

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Answer:

Step-by-step explanation:

Good evening ,

I guess it’s the C expression,

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