In the Cr3+ ion,how many electrons have n+1=4 and ms=-1/2?

Well, I count four, but if that's not your question, you'll have to specify...

Are you asking how many electrons in #"Cr"^(3+)# in atomic orbitals with principal quantum number #n+1 = 4# have #m_s = -1/2#?

If so...

#n = 3#

corresponds to the #3d#, #3p#, and #3s# orbitals, and #m_s = -1/2# designates the spin-down electrons.

#"Cr"# has the electron configuration:

#[Ar]3d^5 4s^1#

Take out one #4s# and two #3d# electrons to get #"Cr"^(3+)#:

#[Ar] 3d^3 = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^3#.

BY CONVENTION, the electron in a singly-occupied orbital is spin-up, i.e. has #m_s = +1/2#, so...

#3d#: ZERO electrons with #m_s = -1/2#

#underbrace(ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))_(3d)#

#3p#: THREE electrons with #m_s = -1/2# (and three with #+1/2#)

#underbrace(ul(uarr color(blue)(darr))" "ul(uarr color(blue)(darr))" "ul(uarr color(blue)(darr)))_(3p)#

#3s#: ONE electron with #m_s = -1/2# (and one with #+1/2#)

#underbrace(ul(uarr color(blue)(darr)))_(3s)#

So there are #bb4# total...