probability qn?

Question

A row of 10 shops is to be painted in 3 colours. 2 shops are to be yellow, 3 to be red and 5 to be blue.

Find the number of ways in which the row of shops can be painted given that the first and the last shops in the same row are of the same colour.

Please help. Thanks!!

Answer 1

2y, 3r, and 5b

8!/0!5!3! + 8!/2!1!5! + 8!/2!3!3!
=56 + 3*56 + 560
=56*14
=784

Answer 2

If the first and the last are yellow, the others are red or blue.

You must choose 3 among 8 for the red shops, then the blue colours are automatically placed and so, there is 8C3 = 56 possibilities

If the first and the last are red, the others are yellow or blue.

You must place the 2 yellow, so 8C2 = 28, the red not placed so 6, and the blue are placed : 6 * 28 = 168 possibilities

It' s just as if the first and the last are blue : 8C2 (yellow) * 6C3 (red) = 28 * 20 = 560
or 8C3 (red) * 5C2(yellow) = 56 * 10 = 560

The numbers of ways is 56 + 168 + 560 = 784

Answer 3

consider all possibility
and add all those things


OK thanks