How many 5-digit numbers divisible by 11 are there containing each of the digits 2,3,4,5,6?

The sum of digits = 2 + 3+ 4 + 5 + 6 = 20

now for the number to be divisible by 11 the sum of digits at odd place should be same as sum of digits at even place or difference multiple of 11

say o is sum od digits at odd place and e sum of digits at even place.

o + e = 20
o- e = 11n and |o - e| < 20 as o+ e = 20

o-e should be even hence should be 0

so e = o = 10
sum of 2 digits ( even placed) and 4 + 6 (only combination)

so 4 and 6 can go to even place

2,3,5 should go to odd place

even position digits can be permuted( 2 ways)  among them selves and odd placed as well ( 6 ways)

so number of numbers = 2 * 6 = 12