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Oliga [24]
3 years ago
12

You are downsizing. Your new residence will have much less closet space than does your current home. You have 12 pair of long pa

nts, 6 pair of short pants and 17 t-shirts. You plan to give some away.
a. In how many ways can a random selection of 9 pair of long pants, 4 pair of short pants and 13 t-shirts be arranged?
b. You randomly select 25 of the items of clothing to give away. What is the probability you select 10 pair of long pants, 3 pair of short pants and 12 t-shirts?
Mathematics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

a) there are 7,854,000 number of ways the random selections can be arranged

b) the required probability is 0.0445

Step-by-step explanation:

Given that;

you have 12 pair of long pants, 6 pair of short pant and 17 t-shirt which in total;

Total = 12 + 6 + 17 = 35

a) number of ways a random selection of 9 pair of long pants, 4 pair of short pants and 13 t-shirts can be arranged;

⇒ [ 12   ×    [ 6     ×    [ 17

     9 ]          4 ]           13 ]

= (12! / (9! 3!)) (6! / (4! 2!)) (17! / (13! 4!))

= (220) (15) (2380)

= 7,854,000

Therefore there are 7,854,000 number of ways the random selections can be arranged

b)

Number of ways of selecting 25 item from 35 items;

⇒ [ 35        

    25 ]

=

(35! / (25! 10!) = 183.579,396

now number of ways to select 10 pair of long pants, 3 pairs of short pantts and 12 t-shirt;

⇒ [ 12   ×    [ 6     ×    [ 17

     10 ]         3 ]           12 ]

= (12! / (10! 2!)) (6! / (3! 3!)) (17! / (12! 5!))

= (66) (20) (6188)

= 8,168,160

So, Required probability = 8,168,160 / 183,579,396

= 0.0445

So the required probability is 0.0445

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