1. Find the number of different ways in which 5 women and 6 men can stand in a line if 4 or more women stand next to one another.
My approach: add no. of ways where exactly 4 women stand next to one another + no. of ways where exactly 5 women stand next to one another
= 7C2 x 2! x 4! x 6! + 7C1 x 5! x 6!
But the answer given is 4233600. Can someone tell me where I went wrong?
2. Find how many 4-digit numbers can be formed using only the digits 3,4,6 and 8, no digit being repeated and if the numbers are
i) divisible by 2
ii) divisible by 4
answer to i) is 18 and ii) is 10. I have i) but I don't know how to get ii) - I don't think it's a P and C problem haha, can someone tell me how you get a number that is divisible by 4 in this case? Does the number have to end with a four etc? Thanks.
i) is 3C1 x 3! = 18, but I don't know how to get ii).
1. Where does the 7C2 x 2! x 4! x 6! come from? To get exactly 4 women next to each other, you have either they are in the middle, in which case you want two men either side, or they are at one end of the line, in which case next to these 4 women must be a man, so ... Or think about inclusion-exclusion. And as a further reason why your answer must be wrong, note that the answer must be divisible by 5!6! (permuting the men, and permuting the women, separately, must give different permutation of the line).
2. Such a number is divisible by 4 if and only if it ends in 36, 48, 64, 68 or 84 (recall divisibility by 4 is determined by the last 2 decimal digits).