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vazorg [7]
3 years ago
10

Exercise topic: Permutations and Combinations. A company wants to hire 3 new employees, but there are 8 candidates, 6 of them wh

ich are men and 2 are women. If the selection is random: a) In how many different ways can choose new employees? b) In how many different ways can choose a single male candidate? c) In how many different ways can choose at least one male candidate? with procedures. Help me please..
Mathematics
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

(a) 56 ways

(b) 6 ways

(c) 56 ways

Step-by-step explanation:

Given:

candidates: 6 mail, 1 female

number to hire : 3

a) In how many different ways can choose new employees?

use the combination formula to choose r to hire out of n candidates

C(n,r) = C(8,3) = 8! / (3! (8-3)! ) = 40320 / (120*6) = 56 ways

b) In how many different ways can choose a single male candidate?

6 ways to choose a male, one way to choose two female, so 6*1 = 6 ways

c) In how many different ways can choose at least one male candidate?

To choose at least 1 male candidate, we subtract the ways to choose no male candidates out of 56.

Since there are only two females, there is no way to choose 3 female candidates.

In other words, there are 56-0 = 56 ways (as in part (a) ) to hire 3 employees with at least one male candidate.

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5x + 2y = -10<br> 3x + y = 66<br><br> please show work
liq [111]

Answer:

x=142, y=-360. (142, -360).

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5x+2y=-10

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----------------

5x+2y=-10

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-----------------------

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-------------------

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