Is Arrangement Enough?
Three ladies have each brought a child for admission to a school. The head of the school wishes to interview the six people one by one, taking care that no child is interviewed before his mother. In how many ways can the interviews be arranged?
The answer is 90.
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1 solution
Nice solution. How I thought of it was to count the number of ways of arranging the letters a , a , b , b , c , c , where the letters correspond to each mother/child pair. Since the mother/child interview order is fixed, there will be a one-to-one correspondence between these ( 2 ! ) 3 6 ! = 8 7 2 0 = 9 0 arrangements and the possible interview orders.
So in general, for n child/mother pairs there are 2 n ( 2 n ) ! possible interview orders, which can be shown to be equivalent to n ! ∗ ( 2 n − 1 ) ! ! .
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Elegant solution indeed! Once again, you can see that I'm a rookie when it comes to combinatorics ;)
Well my answer is 198,which is totally wrong. But I'm not getting where I went wrong. So here's what I did.
3c1 for selecting a mother, and then
You can either select 2 mother or 1 child, so again for second choice 3c1
But for third, if you have selected mother, then you have three choices, either one more mother or any two respective children. So 3c1 OR 2c1
If you've slected mother then in 3! you can selected rest all children,
BUT you've selected child then in 1c1 way you have to first select a mother and then in remaining it will 2! For selecting children.
So here what the total comes. 3 x 3 x 3 x 3 x 2 x 1 + 3 x 3 x 2 x 1 x 2 x 1 = 198
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you are wrong from the 2nd step itself.. (M represents mother and C represents child) (1) M M M C C C 3 × 2 × 1 × 3 × 2 × 1 = 36 (2) M M C M C C 3 × 2 × 2 × 1 × 2 × 1 = 24 (3) M M C C M C 3 × 2 × 2 × 1 × 1 × 1 = 12 (4) M C M C M C 3 × 1 × 2 × 1 × 1 × 1 = 6 (5) M C M M C C 3 × 1 × 2 × 1 × 2 × 1 = 12 Adding 'em you get 90. Hope this helps.
Imagine you are the secretary scheduling the interviews.
First you schedule the mothers; you have 3! = 6 possible orders. You label the mothers A, B, C, in the order of the interviews.
Child c of mother C must be scheduled at the end: A, B, C, c. You can "pencil in" child b in three ways, after B, C, or c. Finally, you have 5 ways to pencil in child a, anytime after mother A, for a grand total of 6 ∗ 1 ∗ 3 ∗ 5 = ( 3 ! ) ∗ ( 5 ! ! ) = 9 0 .
In general, for n mother-child pairs, the interviews can be arranged in ( n ! ) ∗ ( 2 n − 1 ) ! ! ways.