Question

In how many orders can seven different pictures be hung in a row so that one specified picture is at either end?​

Answer 1

Answer:

Step-by-step explanation:

this is a very nice question

lets take the pictures to be 1, 2, 3, 4, 5, 6, 7,

and it is specific for  1 to be hung at either of the ends.

now their are seven places to be filled

_ _ _ _ _ _ _

CASE 1) The first place must have 1, then the second place will have 6 options to fill, the  third will have 5, the fourth will have 4..... until the last place has 1 option

so the no of orders = product of all the number of options

No. of Orders = 720 for case 1

CASE 2) the last place must have one, again, then the first place will have 6 options to fill, the second will have 5, the third will have 4..... until the sixth place has 1 option (as 7th place is already occupied)

No of order= 720 for case 2

total order = sum of both cases = 1440

HOPE THIS HELPS ;)

CHEERS MATE, PLEASE MARK BRAINLIEST

Answer 2

720 possible orders.

Step-by-step explanation:

Let's number the pictures given as 1, 2, 3, 4, 5, 6, and 7.

It needs to be hung such that 1 is at either of the ends.

So the required order is 1, a, b, c, d, e, 1.

There are five places to be filled.

The options for position a will be 2, 3, 4, 5, 6 or 7 = 6 options.

Options for position b will be one less = 5 options

Options for position c will be one lesser = 4 options

Options for position d will be one lesser = 3 options

Options for position e will be one lesser = 2 options.

So total number of options for the order of pictures = 6 * 5 * 4 * 3 * 2 = 720 possible orders.