+0

# Help probability

0
240
6

There is a row of Pascal's triangle that has three consecutive entries,a,b  and c  such that b=2c and a=3c If this row begins 1,n  then find n.

Jul 23, 2022

#4
+118629
+3

Each row of pascal's triangle is symmetrical about the centre so the row that has   3c, 2c, c  also has  c, 2c, 3c

I think this will be easier to work with

$$\binom{n}{r}=c\qquad \binom{n}{r+1}=2c\qquad \binom{n}{r+2}=3c$$

$$\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\ \binom{n}{r+1}=\frac{n!}{r!*(r+1)\;\;*\frac{(n-r)!}{n-r}}=2\binom{n}{r}\\~\\ \qquad \frac{n!}{r!\;\;*(n-r)!}*\frac{(n-r)}{r+1}=2\binom{n}{r}\\~\\ \qquad \binom{n}{r}*\frac{(n-r)}{r+1}=2\binom{n}{r}\\~\\ \qquad \frac{(n-r)}{r+1}=2\\~\\ \qquad (n-r)=2r+2\\~\\ \qquad n=3r+2\\~\\$$

$$\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\ \binom{n}{r+2}=\frac{n!}{r!*(r+1)(r+2)\;\;*\frac{(n-r)!}{(n-r)(n-r-1)}}=3\binom{n}{r}\\~\\ \qquad \frac{n!}{r!\;\;*(n-r)!}*\frac{(n-r)(n-r-1)}{(r+1)(r+2)}=3\binom{n}{r}\\~\\ \qquad \binom{n}{r}*\frac{(n-r)(n-r-1)}{(r+1)(r+2)}=3\binom{n}{r}\\~\\ \qquad \frac{(3r+2-r)(3r+2-r-1)}{(r+1)(r+2)}=3\\~\\ \qquad \frac{(2r+2)(2r+1)}{(r+1)(r+2)}=3\\~\\ \qquad 4r^2+6r+2=3(r^2+3r+2)\\~\\ \qquad r^2-3r-4=0\\ \qquad(r-4)(r+1)=0\\ \qquad \text{r is positive so r=4}\\ n=3r+2\\ so \\ n=14\\ check\\ \binom{14}{4}=1001\\ \binom{14}{5}=2002\\ \binom{14}{6}=3003$$

LaTex:

\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\
\binom{n}{r+1}=\frac{n!}{r!*(r+1)\;\;*\frac{(n-r)!}{n-r}}=2\binom{n}{r}\\~\\

\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\
\binom{n}{r+2}=\frac{n!}{r!*(r+1)(r+2)\;\;*\frac{(n-r)!}{(n-r)(n-r-1)}}=3\binom{n}{r}\\~\\
\qquad  \text{r is positive so r=4}\\
n=3r+2\\
so \\
n=14\\
check\\
\binom{14}{4}=1001\\
\binom{14}{5}=2002\\
\binom{14}{6}=3003

Jul 23, 2022

#1
0

Jul 23, 2022
#2
0

$$n = \boxed{10}$$

.
Jul 23, 2022
#3
0

Sorry that i incorrect

Guest Jul 23, 2022
#4
+118629
+3

Each row of pascal's triangle is symmetrical about the centre so the row that has   3c, 2c, c  also has  c, 2c, 3c

I think this will be easier to work with

$$\binom{n}{r}=c\qquad \binom{n}{r+1}=2c\qquad \binom{n}{r+2}=3c$$

$$\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\ \binom{n}{r+1}=\frac{n!}{r!*(r+1)\;\;*\frac{(n-r)!}{n-r}}=2\binom{n}{r}\\~\\ \qquad \frac{n!}{r!\;\;*(n-r)!}*\frac{(n-r)}{r+1}=2\binom{n}{r}\\~\\ \qquad \binom{n}{r}*\frac{(n-r)}{r+1}=2\binom{n}{r}\\~\\ \qquad \frac{(n-r)}{r+1}=2\\~\\ \qquad (n-r)=2r+2\\~\\ \qquad n=3r+2\\~\\$$

$$\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\ \binom{n}{r+2}=\frac{n!}{r!*(r+1)(r+2)\;\;*\frac{(n-r)!}{(n-r)(n-r-1)}}=3\binom{n}{r}\\~\\ \qquad \frac{n!}{r!\;\;*(n-r)!}*\frac{(n-r)(n-r-1)}{(r+1)(r+2)}=3\binom{n}{r}\\~\\ \qquad \binom{n}{r}*\frac{(n-r)(n-r-1)}{(r+1)(r+2)}=3\binom{n}{r}\\~\\ \qquad \frac{(3r+2-r)(3r+2-r-1)}{(r+1)(r+2)}=3\\~\\ \qquad \frac{(2r+2)(2r+1)}{(r+1)(r+2)}=3\\~\\ \qquad 4r^2+6r+2=3(r^2+3r+2)\\~\\ \qquad r^2-3r-4=0\\ \qquad(r-4)(r+1)=0\\ \qquad \text{r is positive so r=4}\\ n=3r+2\\ so \\ n=14\\ check\\ \binom{14}{4}=1001\\ \binom{14}{5}=2002\\ \binom{14}{6}=3003$$

LaTex:

\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\
\binom{n}{r+1}=\frac{n!}{r!*(r+1)\;\;*\frac{(n-r)!}{n-r}}=2\binom{n}{r}\\~\\

\binom{n}{r}=\frac{n!}{r!*(n-r)!}\\~\\
\binom{n}{r+2}=\frac{n!}{r!*(r+1)(r+2)\;\;*\frac{(n-r)!}{(n-r)(n-r-1)}}=3\binom{n}{r}\\~\\
\qquad  \text{r is positive so r=4}\\
n=3r+2\\
so \\
n=14\\
check\\
\binom{14}{4}=1001\\
\binom{14}{5}=2002\\
\binom{14}{6}=3003

Melody Jul 23, 2022
#5
+129799
+1

Very nice, Melody    !!!!!

CPhill  Jul 23, 2022
#6
+118629
0

Thanks Chris

Melody  Jul 24, 2022