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3 years ago

Need help with this shape

Mathematics

1 answer:

NISA [10]3 years ago

4 0

It is a quadrilateral

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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Solved

6 0

2 years ago

Simplify the expression -1/8-2/7

SpyIntel [72]

the answer would be $-\frac{23}{56}$ because you would need to find the LCD, which in this case would be 56 and you solve from there.

4 0

3 years ago

Read 2 more answers

SOMEONE PLEASE HELP!!

german

$-12=\dfrac{-12}{1}=\dfrac{12}{-1}=\dfrac{-24}{2}=\dfrac{24}{-2}=\dfrac{-36}{3}=\dfrac{36}{-3}=\dfrac{-48}{4}=\dfrac{48}{-4}=\dots$

7 0

3 years ago

What is the word form for 2.51 and expanded form

Nimfa-mama [501]

Word Form:
Two and fifty-one hundredths

Expanded Form:
2 + 0.5 + 0.01

Hope I Helped You!!! :-)

Have A Good Day!!!

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