Can mode lie between mean and median? The distribution is left-skewed if mean<median<mode.
The distribution is right-skewed if mean>median>mode.
Can mode lie between mean and median?
 A: First, left- or right-skewed tells you nothing about the mode.
For a case you are looking for try out
$$0, 1, 1.5, 1.6, 2, 2, 12$$
A: Paul's answer works, but it is also possible to have a distribution which is unimodal and either is continuous or is discrete with equal gaps in the support, such as the following distribution with median $2$, mode $3$ and mean $3.02$:
x   P(X=x)
0   0.16
1   0.17
2   0.18
3   0.19
4   0.06
5   0.055
6   0.05
7   0.045
8   0.04
9   0.03
10  0.02

It is possible to have the mean, median and mode in any order, though having the mode strictly between the other two may in a vague handwaving sense be less likely than other patterns, as I once tried to show
A: Yes, The first two sentences are often, but not always, true. (They are true for some
well-known distributions.)
However, as you can see from other Answers,
and from my example below, it is possible to contrive a sample or population
with discrete values, for which the mode lies between the median and the mean. [Using R.]
x = c(1,1,1,2,2,3,3,3,4,4,4,4,25)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 1.000   2.000   3.000   4.385   4.000  25.000 

Median is $3,$ mode is $4,$ mean is $4.385.$
Exponential distribution. Among frequently-used distributions, consider right-skewed $\mathsf{Exp(1)}.$
Mode is $0$, median is $0.6931$ (below), and mean is $1.$
qexp(.5, 1)
[1] 0.6931472

A beta distribution. For the left-skewed continuous distribution $\mathsf{Beta}(5,2),$
the mean is $5/7 \approx 0.714,$ the median is $0.736,$
and the mode is $4/5 = 0.800.$
qbeta(.5, 5, 2)
[1] 0.73555

Samples from continuous distributions. It is not so easy to define the mode of a sample from
a continuous distribution because no two sample values
(before rounding) can ever be exactly the same.
Two (of many) possible definitions of the mode of the beta sample below
might be the center of the tallest bar of its histogram
(which is somewhat arbitrary depending on binning), or
the maximum of its kernel density estimator (KDE), which we can
view for the moment as a 'smoothed ideal histogram'.
set.seed(2022)
x = rbeta(1000, 5, 2)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.1800  0.5991  0.7428  0.7095  0.8381  0.9976 
hist(x, prob=T, ylim=c(0,3), col="skyblue2")
lines(density(x), col="orange", lwd=2)


So, the sample mean is $0.710,$ the sample median is $0.742,$
and the sample mode might be taken to be something like $0.75$ (from the tallest histogram bar) or something like $0.79$ from the KDE (orange curve).
A: In case of a positively skewed frequency distribution, the mean is always greater than median and the median is always greater than the mode. In case of a negatively skewed frequency distribution, the mean is always lesser than median and the median is always lesser than the mode.
