Maths-
General
Easy

Question

The number of ways in which 17 billiard balls be arranged in a row if 7 of them are black, 6 are red, 4 are white is

  1. fraction numerator 10 factorial over denominator 6 factorial 4 factorial end fraction    
  2. 17 factorial    
  3. fraction numerator 17 factorial over denominator 6 factorial 7 factorial 4 factorial end fraction    
  4. 6 factorial 7 factorial 4 factorial    

Hint:

To solve this question, we will assume that all the balls of the same color are identical. Now, to arrange all these balls, we will apply the following formula:
T o t a l space A r r a n g e m e n t s space equals fraction numerator X factorial over denominator left parenthesis x 1 factorial right parenthesis left parenthesis x 2 factorial right parenthesis left parenthesis x 3 factorial right parenthesis...... left parenthesis x n factorial right parenthesis end fraction

The correct answer is: fraction numerator 17 factorial over denominator 6 factorial 7 factorial 4 factorial end fraction


    Detailed Solution
    I n space t h e space a b o v e space f o r m u l a comma space X space equals space t o t a l space n u m b e r space o f space e n t i t i e s comma space x subscript 1 space space a r e space t h e space s a m e space k i n d space o f space e n t i t i e s comma space t h a t space i s space t h e y space a r e space i d e n t i c a l comma space x subscript 2 space space a r e space a l s o space t h e space s a m e space k i n d space o f space e n t i t i e s space b u t space d i f f e r e n t space f r o m space x subscript 1 space. space

S i m i l a r l y comma space x subscript 3 comma x subscript 4 comma x subscript 5..... x subscript n space space a r e space t h e space n u m b e r space o f space i d e n t i c a l space e n t i t i e s space b u t space d i f f e r e n t space f r o m space e a c h space o t h e r.
    Before solving the question, we are going to assume that all the balls of the same color are identical. This means that one red ball is identical to another red ball. Similarly, one black ball is identical to all the other black balls and one white ball will be similar to other white balls. Now, we are given that out of 17 balls, 7 of them are black, 6 are red and 4 are white. Now, we will arrange these balls in a row. The formula by which we can arrange the total number of entities which contain similar entities is given as:
    T o t a l space A r r a n g e m e n t s space space equals space fraction numerator X factorial over denominator left parenthesis x subscript 1 factorial right parenthesis left parenthesis x subscript 2 factorial right parenthesis left parenthesis x subscript 3 factorial right parenthesis...... left parenthesis x subscript n factorial right parenthesis end fraction

I n space t h i s space f o r m u l a comma space X space i s space t h e space t o t a l space n u m b e r space o f space e n t i t i e s comma space x subscript 1 space space i s space t h e space n u m b e r space o f space i d e n t i c a l space e n t i t i e s space o f space t h e space f i r s t space k i n d comma space x subscript 2 space space i s space t h e space n u m b e r space o f space i d e n t i c a l space e n t i t i e s space o f space t h e space s e c o n d space k i n d comma space a n d space s o space o n. space

I n space o u r space c a s e comma space X equals 17 space comma x subscript 1 equals 7 space comma x subscript 2 equals 6 space space a n d space x subscript 3 equals 4 space.
    Thus, we get,
    T o t a l space A r r a n g e m e n t s space equals space fraction numerator 17 factorial over denominator left parenthesis 7 factorial right parenthesis left parenthesis 6 factorial right parenthesis left parenthesis 4 factorial right parenthesis end fraction space equals space bold 17 bold cross times bold 14 bold cross times bold 13 bold cross times bold 11 bold cross times bold 10 bold cross times bold 3 bold cross times bold 4 bold space bold space bold equals bold 4084080 bold space bold space
    Thus, there are 4084080 ways in which we can arrange these billiard balls.

    We cannot arrange the billiards balls as follows:
    There are 17 balls, so the total number of arrangements = 17!
    This is incorrect because this method is applicable only when the balls are distinct, not identical. But, in our case, all the balls of the same color are identical.

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