00:01
Hi, from the question given that, let as a sequel to set contains 1, 2, 3, 4, 6.
00:12
So in part a, we need to find how many subsets are there of cardinality 4.
00:18
So, cardinality 4.
00:25
So here, total number of elements is 6.
00:28
Out of 6 we need to choose 4 elements.
00:31
So there will be 6c4 ways.
00:38
To select four elements from six.
00:54
Therefore, n -c -r can be written as n -factoral divided by r -factoril times n -minus r -factoral.
01:03
So here, 6 -4 will be equal to 6 -factoral divided by 4 -factoral, 6 -minus 4 -4 became 2 -factoral.
01:12
So this can be written as 6 times 5 times 4 -factoral divided by 4 -factoral.
01:18
Times 2 factor l so this and this will get cancelled and 2 factorial we have 3 so totally 15 base hence we conclude that there will be 15 base so these are the subset of cardinality 4 so by using this we need to find how many subsets having cordonality 4 having the set 2, 3, 5.
02:06
So, from this, this one contained 2, 3, 5 and this.
02:19
So totally number of subset of cardinality 4 contains subset 235 will be 3...