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4 mathematics and 5 physics books are arranged in a row on a shelf at random. Find the probability that the books on the same subject will be put together.
A. 1/2520
B. 1/1260
C. 1/126
D. 1/63
E. 2/63
_________________
A. 1/2520
B. 1/1260
C. 1/126
D. 1/63
E. 2/63
_________________
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Archit3110
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Bunuel
Proabilty would be :
4!*5!*2/9!
4*3*2*1*2/(9*8*7*6)
=1/63
IMO D
komals06
Joined: 10 Nov 2014
Last visit: 20 Jul 2023
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Location: United States
Concentration: Marketing, Strategy
Schools: Goizueta '22 IIMA PGPX "21
Schools: Goizueta '22 IIMA PGPX "21
Posts: 27
Soln:
( Given)There are - 4 Maths books- say m1,m2,m3, and 5 physics books-- say p1, p2, p3,p4,p5
Total - 9 books
We need to find Probability (The books on the same subject will be put together)
Total ways we can arrange 9 books =9! ways
Fav outcome= 4 maths books can be arranged in 4! ways
5 physics books can be arranged in 5! ways within themselves
and considering Maths books and physics books are two groups. Two groups [m1,m2,m3,m4] and [p1,p2,p3,p4,p5] can arrange themselves in 2! ways
Therefore - Favourable outcome = 5!*4!*2! ( no of ways each group arrange within same category * no of ways each grp arrange with eachother)
Already we know Total outcome =9! ( 9 books arranging without any constraint)
Probability = fav outcome/ total outcome
Probability = 5!*4!*2!/ 9!
=1/63
Ans choice D
( Given)There are - 4 Maths books- say m1,m2,m3, and 5 physics books-- say p1, p2, p3,p4,p5
Total - 9 books
We need to find Probability (The books on the same subject will be put together)
Total ways we can arrange 9 books =9! ways
Fav outcome= 4 maths books can be arranged in 4! ways
5 physics books can be arranged in 5! ways within themselves
and considering Maths books and physics books are two groups. Two groups [m1,m2,m3,m4] and [p1,p2,p3,p4,p5] can arrange themselves in 2! ways
Therefore - Favourable outcome = 5!*4!*2! ( no of ways each group arrange within same category * no of ways each grp arrange with eachother)
Already we know Total outcome =9! ( 9 books arranging without any constraint)
Probability = fav outcome/ total outcome
Probability = 5!*4!*2!/ 9!
=1/63
Ans choice D
Posts: 410
4M and 5P and we are to treat each individual book in each category the same i.e. p1=p2
There are a total of 9!/5!4! ways to arrange the bookshelf – there are 9! ways to arrange 9 different items, but we must reduce that number as P and M are repeated
There are two arrangements that allows all M and all P to be together: MMMMPPPPP and PPPPPMMMM
\(=\frac{2}{\frac{9!}{5!4!}}\)
\(=\frac{2}{\frac{9·8·7·6·5!}{5!4!}}\)
Cancel the common 5!
\(=\frac{2}{\frac{9·8·7·6}{4·3·2·1}}\)
\(=\frac{2}{\frac{9·7·2}{1}}\)
\(=\frac{2}{9·7·2}\)
\(=\frac{1}{63}\)
There are a total of 9!/5!4! ways to arrange the bookshelf – there are 9! ways to arrange 9 different items, but we must reduce that number as P and M are repeated
There are two arrangements that allows all M and all P to be together: MMMMPPPPP and PPPPPMMMM
\(=\frac{2}{\frac{9!}{5!4!}}\)
\(=\frac{2}{\frac{9·8·7·6·5!}{5!4!}}\)
Cancel the common 5!
\(=\frac{2}{\frac{9·8·7·6}{4·3·2·1}}\)
\(=\frac{2}{\frac{9·7·2}{1}}\)
\(=\frac{2}{9·7·2}\)
\(=\frac{1}{63}\)
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