How many ways 10 hats of different colours can be put on the heads of 10 boys sitting in a row, so that the red and the black coloured hats can never be put on the heads of any two adjacent boys? someone plz explain and solve ....plz
Answers
Answer:
The answer would be 10! if there were no restrictions.
If the hats were:
A B C D E F G H I J
and there were no restrictions they could be arranged in 10! ways
From that 10! we must subtract the number of non-allowable ways.
Suppose the red hat is hat A and the black hat is hat B.
Step-by-step explanation:
Then we must subtract the 9! ways they could be
arranged with hat A and hat B together, with hat A
left of B:
AB C D E F G H I J
And we must also subtract the 9! ways they could be
arranged with hat A and hat B together, with hat B left
of hat A:
BA C D E F G H I J
Answer: 10! - 2*9! = 3628800 - 2*(362880) = 2903040
Edwin
8 * 9! ways such that red and black coloured hats can never be put on the heads of any two adjacent boys
Step-by-step explanation:
10 hats of different colours on the heads of 10 boys sitting in a row
can be put in 10! ways
Now let say red & Black hat are put on two adjacent boys head
Two adjacent boys can be selected in 9 ways
red & hat cap can be placed in 2 ways
remaining 8 boys with 8 hats can be done in 8! ways
number of Ways = 9 * 2 * 8!
= 2 * 9!
Number of ways so that the red and the black coloured hats can never be put on the heads of any two adjacent boys = 10! - 2 * 9!
= 10 * 9! - 2 * 9!
= 8 * 9!
8 * 9! ways
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