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In the questions below nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture. In how many ways can this be done if a) Ben is to be in the picture? b) Both Ed and Gail are in the picture? 4200 c) Neither Ed nor Fran are in the picture? 5040 d) Dot is on the left end and Ed is on the right end? 210 e) Hal or Ida (but not both) are in the picture? 4200 f) Ed and Gail are in the picture, standing next to each other? 960 g) Ann and Ben are in the picture, but not standing next to each other?

User One Lyner
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5 votes

Answer:

Explanation:

Ben is present in all pictures so rest of 4 persons to be selected from 8 persons , no of ways to do it is

⁸C₄ = 8 x 7 x 6 x 5 / 4 x 3 x 2 x 1

= 70 .

In one of these combination , we can get 5 ! ( five factorial )

Total no of permutations

= 70 x 5 !

= 8400 .

b ) Both Ed and Gail are present

The above derivation changes to the following .

Total no of permutations

= ⁷C₃ x 5 !

= 35 x 120

= 4200

c )

exclude 2 person from 9 to be selected

permutation

= ⁷C₅ x 5 !

= 21 x 120

= 2520

d )

rest of 3 person from 7 persons , no of permutations

⁷P₃ = 7 x 6 x 5 = 210

e )

Hal or Ida can occupy any of the 5 position which can be done in 5 ways .

when one position is occupied , the rest of 4 position can be occupied by 7 persons which can be done in

⁷P₄ ways

Total ways = 5 x ⁷P₄

= 5 x 840

= 4200

f )

They can occupy position like 1,2 or 2,3 or 3,4 or 4,5

Rest of the position can be occupied in ⁷P₃ ways

Total ways = 4 x ⁷P₃

= 4 x 210

= 840

They can also be exchanged mutually so no of ways

= 840 x 2 = 1680 .

g ) No of pictures in which Ann and Ben are present

two position to be selected out of 5 = ⁵P₂

Rest of position that can be shuffled = ⁷P₃

Total no of pictures in which both are present

= ⁵P₂ x ⁷P₃

= 20 x 210

= 4200

out of which they will be standing next to each other = 1680

no of pictures in which they will not be standing next to each other

= 4200 - 1680 = 2520. .

User Houssam Hamdan
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