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Anvisha [2.4K]
4 years ago
10

When selecting subsets of a larger set in which order is not important, you are finding the number of ________ of n elements tak

en r at a time.
Fill in the blanks.
Mathematics
1 answer:
Scorpion4ik [409]4 years ago
5 0

Answer:

When selecting subsets of a larger set in which order is not important, you are finding the number of combination of n elements taken r at a time.

Step-by-step explanation:

We have to fill the blank.

When selecting subsets of a larger set in which order is not important, you are finding the number of combination of n elements taken r at a time.

Combination:

  • A combination is a selection of all or part of a set of objects from a larger set such that the order of selection is not important.
  • This is given by the formula:

C(n,r) = ^nC_r = \dfrac{n!}{r!(n-r)!}

where n is the number of objects in consideration or the larger set from which we have to choose r with no repetition.

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Step-by-step explanation:

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6 0
3 years ago
Calculate the sum of the multiples of 4 from 0 to 1000
allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d  can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 =   (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

       = (2*0 + (251-1)4)251/2

       = (250*4)251/2

     = 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

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