Asked by sh
What is the third number in the 156th row of Pascal's triangle?
I know that it is 154 x ___
how do I find out the number?
thanks in advance
I know that it is 154 x ___
how do I find out the number?
thanks in advance
Answers
Answered by
Reiny
let's look at the triangle for a few rows
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
let's also agree that, say, we use
row 6, that would be 1 5 10 10 5 1
notice that C(5,2) which is 10 gives us the 3rd term in row 6
so the third term in row 156 is
C(155,2)
= 155!/(2!153!) = 155(154)/2 = 11935
another way is to realize that the third column in Pascal's triangle makes up the "triangular" numbers.
(if you play billiards, think of the triangle of balls at the start, in the first row there is one ball, including the first and the second row you have 3 balls, including the first, second and third row you have 6 balls et.)
these numbers are generated by (n-1)(n-2)/2 for the third term in the nth row
so for the third term in row 156 we would have 155*154/2 just like above.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
let's also agree that, say, we use
row 6, that would be 1 5 10 10 5 1
notice that C(5,2) which is 10 gives us the 3rd term in row 6
so the third term in row 156 is
C(155,2)
= 155!/(2!153!) = 155(154)/2 = 11935
another way is to realize that the third column in Pascal's triangle makes up the "triangular" numbers.
(if you play billiards, think of the triangle of balls at the start, in the first row there is one ball, including the first and the second row you have 3 balls, including the first, second and third row you have 6 balls et.)
these numbers are generated by (n-1)(n-2)/2 for the third term in the nth row
so for the third term in row 156 we would have 155*154/2 just like above.
Answered by
sh
hmm.. 155!/(2!)(153!) doesn't work on the calculator.
So the fourth term would be 155x154x153/3?
So the fourth term would be 155x154x153/3?
Answered by
Reiny
you are right, most calculators cannot go over 69!
but if you break down 155!/(2!153!)
by its definition:
155! = (155)(154)153!
then 155!/(2!153!)
= (155)(154)153! / (2!153!)
= 155(154)/2
look for a key something like <sub>n</sub>C<sub>r</sub>
by definition that is n!/(r!(n-r)!)
<sub>n</sub>C<sub>r</sub> is just another notation for C(n,r)
with that key, your calculator should be able to do C(155,2)
on my calc I do :
155
2nd F
the <sub>n</sub>C<sub>r</sub> , (under the 5 on mine)
2
=
to get 11935
but if you break down 155!/(2!153!)
by its definition:
155! = (155)(154)153!
then 155!/(2!153!)
= (155)(154)153! / (2!153!)
= 155(154)/2
look for a key something like <sub>n</sub>C<sub>r</sub>
by definition that is n!/(r!(n-r)!)
<sub>n</sub>C<sub>r</sub> is just another notation for C(n,r)
with that key, your calculator should be able to do C(155,2)
on my calc I do :
155
2nd F
the <sub>n</sub>C<sub>r</sub> , (under the 5 on mine)
2
=
to get 11935
Answered by
sh
Yep, I learned the C button already, thank you very much for the explanation:)
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