Pascal triangles are triangle of numbers formed by binomial coefficient. There are only two entries that equals to 41.
Given that:
[tex]Entry= 41[/tex]
The entries of pascal triangle are represented by using combination formula.
[tex]^nC_r = \frac{n!}{(n -r)!r!}[/tex]
We want to find the number of equations where the end result equals 41.
i.e. [tex]^nC_r = 41[/tex]
This gives:
[tex]\frac{n!}{(n -r)!r!} = 41[/tex]
Rewrite as:
[tex]\frac{n!}{(n -r)!r!} = \frac{41\times 40!}{40!}[/tex] ---- this is true because [tex]\frac{40!}{40!} =1[/tex]
Multiply the denominator by 1!
[tex]\frac{n!}{(n -r)!r!} = \frac{41 \times 40!}{40! \times 1!}[/tex]
The numerator becomes
[tex]\frac{n!}{(n -r)!r!} = \frac{41!}{40! \times 1!}[/tex]
The above implies:
[tex]^nC_r =\frac{41!}{40! \times 1!}[/tex]
The value of r can take any of the factor in the denominator (without the factorial).
So:
[tex]^nC_r =^{41}C_{40}[/tex] or [tex]^nC_r =^{41}C_{1}[/tex]
Hence, there are two entries of the Pascal triangle that equal 41
Learn more about Pascal triangle at:
https://brainly.com/question/12668060
