1The 5 people who row can be selected in 10C5 ways. The 5 can be arranged in 5! ways. The rest can be arranged in 25 ways.
So is it - 10C5 * 5! * 25 ?
A boat has a crew of 10 men of which 3 can row only on one side and 2 only on the other.In how many ways,can the crew be arranged??
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9 Answers
62It is not given that there how many guys on each side, so this is not a very easy problem!
What can happen is that there can be any no of guys on either side..
so we have to take these cases.
Case: 3 on left side and 7 on right
No of ways to select=1;
No of ways to arrange=3!.7!
Case: 4 on left side and 6 on right
No of ways to select=5;
No of ways to arrange=4!.6!
Case: 5 on left side and 5 on right
No of ways to select=5C2;
No of ways to arrange=5!.5!
Case: 6 on left side and 4 on right
No of ways to select=5C3;
No of ways to arrange=6!.4!
Case: 7 on left side and 3 on right
No of ways to select=5C4;
No of ways to arrange=7!.3!
Case: 8 on left side and 2 on right
No of ways to select=5C5;
No of ways to arrange=7!.2!
Take the sum of products of the terms in each case.. u get the answer :)
1wen we have 7 on one side...rnt we goin to use combination to choose 3 out of 7 men.......that is 7c3 rit............how can we purmuteas 3!
i think u r mistaken bro.........
for the case 7 on 1 side n 3 on the oder
out of seven 3 can be chossen in 7c3 ways..n out of 3 ..2 can b choosen in 3c2 ways........so it becomes....
10c3 * 3c3 * 7c3 * 3C2 ways....
similar way oder cases should b mede ........rit??
62OOPPSS> i just needed to find the no of ways the crew has to be arranged.. not the no of ways they can be seated :D
see that is far far far far far simpler...
out of the 10 ppl there, 5 are fixed..
the other 5 can be put on either side .. each in 2 ways..
so the total no of ways is 25
162dude u never said that there will be equal men on each side of the boat!
ur answer is for that case :)
3the answer is 5C2 * 5! * 5!
MY LOGIC IS....
leaving out the people who wish to sit according to their desire, we have 10-2-3=5 people left!! now we 1st place the 3+2=5 men according to their wish.....to have 5 men on each side, we can consider only one side and the other will be adjusted accordingly.
Let us consider the side which now has 3 men sitting...2places are vacant...we can select any 2 from the remaing 5 to sit there ....thus we have 5C2
if we consider the other side, we will get 5C3 and we know that....
5C2=5C3!!!
NOW WE CAN RANDOMLY ARRANGE BOTH SIDES IN 5! WAYS!!
THUS WE GET THE ABOVE ANSWE...hope it is correct!! :)
1@ Nishant Sir : In a boat there should be equal number of people on each side...isnt it ?