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Radda [10]
3 years ago
9

A die is rolled 20 times. given that three of the rolls came up 1, five came up 2, four came up 3, two came up 4, three came up

5, and three came up 6, how many different arrangements of the outcomes are there
Mathematics
1 answer:
Serjik [45]3 years ago
8 0

Solution:

Number of times a die is rolled = 20

1 - 3=A

2 - 5=B

3 - 4=C

4 - 2=D

5 - 3=E

6 -  3=F

Total number of arrangements of outcomes , when a dice is rolled 20 times given that 1 appear 3 times, 2 appears 5 times, 3 appear 4 times, 4 appear 2 times , 5 appear three times, and 6 appear 3 times

            = Arrangement of 6 numbers (A,B,C,D,E,F) in 6! ways and then arranging outcomes

= 6! × [ 3! × 5! × 4!×2!×3!×3!]

= 720 × 6×120×24×72→→[Keep in Mind →n!= n (n-1)(n-2)(n-3)........1]

= 895795200  Ways

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Step-by-step explanation:

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3 years ago
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7 0
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If LN = 6x - 5, LM = x + 7, and MN = 3x + 20, find MN.<br> A.68<br> B.16<br> C.23<br> D.18
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Answer : The correct option is (A) 68.

Step-by-step explanation :

As we are given that:

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Now we have to determine the value of MN.

According to the question:

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Now putting all the given values in this expression, we get:

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The value of MN = 3x + 20 = 3(16) + 20 = 48 + 20 = 68

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