Answer:
Step-by-step explanation:
Given that there are six different toys and they are to be distributed to three different children.
The restraint here is each child gets atleast one toy.
Let us consider the situation as this.
Since each child has to get atleast one toy no of ways to distribute
any 3 toys to the three children each. This can be done by selecting 3 toys from 6 in 6C3 ways and distributing in 3! ways
So 3 toys to each one in 6x5x4 =120 ways
Now remaining 3 toys can be given to any child.
Hence remaining 3 toys can be distributed in 3x3x3 =27 ways
Total no of ways
= 120(27)
= 3240
The answer is g=13 I hope this helped
Answer:
DC = 2
Step-by-step explanation:
The wrong equation was used.
The right equation to use based on the midsegment theorem of a trapezoid is:
MN = ½(AB + DC)
MN = 8
AB = 14
Substitute
8 = ½(14 + DC)
Multiply both sides by 2
8*2 = ½(14 + DC)*2
2*8 = 14 + DC
16 = 14 + DC
16 - 14 = 14 + DC - 14
2 = DC
DC = 2
➻ In a group of 40 people, 27 can speak English and 25 can speak Spanish.
➻ The required number of people who can speak both English and Spanish .
<u>Consider</u> ,
➻ A → Set of people who speak English.
➻ B → Set of people who speak Spanish
➻ A∩B → Set of people who can speak both English and Spanish
➻ n(A∪B) = n(A) + n (B) - n(A∩B)
➻ 40 = 27 + 25 - n (A∩B)
➻ 40 = 52 - n (A∩B)
➻ n (A∩B) = 52 - 40
➻ ∴ n (A∩B) = 12
∴ Required Number of persons who can speak both English and Spanish are <u>12 .</u>
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➻ n(A∪B) = n(A) + n (B) - n(A∩B)
➻ 40 = 27 + 25 - 12
➻ 40 = 52 - 12
➻ 40 = 40
➻ ∴ L.H.S = R.H.S
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