Q:)In how many ways can we put 7 different balls into 5 different boxes, such that no box is empty?
Why won't the following approach work?
First, we select 5 balls out of 7 in 7C5 = 21 ways.
These 5 balls are distributed in 5 boxes in 5! = 120 ways, now that 5 balls have been put in 5 boxes, one in each, so that no box remains empty.
The remaining 2 balls are put in the 5 boxes in 5X5 = 25 ways.
=>Ans.= 21X120X25 = 63000.
OA is given as 16800.
Please tell me where my approach is wrong.
Question :
In how many ways 10 identical presents be distributed among 6 children such that none gets zero.
a+b+c+d+e+f = 10
none of them cannot be 1
so , a+b+c+d+e+f = 10-6 = 4
=> 9C5
6 people have to go to attend a meeting in 3 different vehicles, each of which can accommodate 6 persons. In how many ways they can go so that they use all the three vehicles?
Please solve this with clear steps.
1,2,3 = 6C3*3C2*3! = 360
1,1,4 = 6C4*2C1*3!/2 = 90
2,2,2 = 6C2*4C2 = 90
so,total ways = 540
Q:)In how many ways can we put 7 different balls into 5 different boxes, such that no box is empty?
Why won't the following approach work?
First, we select 5 balls out of 7 in 7C5 = 21 ways.
These 5 balls are distributed in 5 boxes in 5! = 120 ways, now that 5 balls have been put in 5 boxes, one in each, so that no box remains empty.
The remaining 2 balls are put in the 5 boxes in 5X5 = 25 ways.
=>Ans.= 21X120X25 = 63000.
OA is given as 16800.
Please tell me where my approach is wrong.
sorry yar i am not getting your approach..try it
1,1,1,1,3 = 7C3*4!*5 = 4200
1,1,1,2,2 = 7C2*5C2*3!*10 = 12600
total = 16800
rajr Sayshow many 4 digit numbers that are divisible by 3,can be formed using the digits 0,1,2,3 and 8 if no digit is to occur more than once in each number?
0+1+2+3+8 = 14
now, we have to form 4 digit number and we have 5 digits
when sum=12
if we won't consider 2 here we will get sum of 12
_ _ _ _ = 3*3*2*1 = 18
when sum=6
_ _ _ _ = 3*3*2*1 = 18
total = 36
Sorry for this silly que bt m nt getting the answer given in the book ...
The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is?
840 or 720?
selecting places for A,B,C can be done in 7C3 i.e. 35 ways and they can be arrange in only 1 way
now other 4 members can be arrange themselves in 4! ways
so,total ways = 35*4! = 840
30. What is the number of ways in which a team of 3 members can be selected from 4 boys and 6 girls such that there is at least one boy and at least one girl in the team?
The answer is (4C1*6C1*8C1)/2.
My question is why the division by 2?
Can you explain the second problem more clearly, I didnt get the 72 part.
I dont know why this question has been left unslolved... So even though u might have got the answer it could be helpfull someone else who is looking out for solution!!
For first
The word INKLING has 5 distinct letters and 2 letters are repeating...
1.U can select 4 letters from 5 distinct letters in 5c4 ways=5
2.When 2 I`s are repeating, u can arrange them in 1x4c2 ways=6(1x5c2 bcoz u can select `I`I` in one way and remaining 2 letters out of 4 distinct letters can be done in 4c2 ways)
3.When 2 N`s are repeated, u can arrange them in 1x4c2 ways=6
4.the combination IINN can be done in 1way... Remember the arrangement (ININ or NINI or NNII or any other arrangement doesn`t matter)
so 5+6+6+1=18 ways
2) Now second is bit tricky
1)arrange 5 distinct letter(INKLG) This can be done in 5 x 4 x 3 x 2 x1 =120
2)arrange II and NKLG, this can be done in 13/2!x4x3=72
(12/2! bcoz I can be arranged in 4 ways and other I can be arranged in 3 ways.. Dividing by two bcoz we might encounter a sitation of I1I2KLor I2I1KL so we dont want such combunation)
3)arrange NN and IKLG this can be done in 72 ways similar to last case
4)arrange II and NN this can be done in 4!/2!2!=6 ways
so total arrangement of 120+72+72+6=270
I hope I got both of them right
Can you please explain the second problem more clearly
1) Find the number of selections that can be made by taking 4 letters from the word INKLING?
2)arrangements for by taking 4 letters for the word above
30. What is the number of ways in which a team of 3 members can be selected from 4 boys and 6 girls such that there is at least one boy and at least one girl in the team?
The answer is (4C1*6C1*8C1)/2.
My question is why the division by 2?
See.. It is like this..
When you have 4 boys and 6 girls, you can pick:
Case 1: One boy and two girls or
Case 2: One girl and two boys
Now, when you pick one boy, one girl; you are left with 8 people. And you have to chose one of them. Say you choose a boy.
Now, first boy can be chosen in 4 ways. Say, B1 was chosen. And then from remaining 8, you chose B2.
Then there will be one case, when we choose B2 as first boy. And from remaining 8; we choose B1. So, we get same case.
So, like this, each case when we calculate like (4C1 * 6C1 * 8C1) will be counted twice.. Hence, divide by 2.
We can also solve it by making two cases:
Case 1: One boy and two girls => 4C1 * 6C2 = 4*15 = 60 ways
Case 2: One girl and two boys => 6C1 * 4C2 = 6*6 = 36 ways
Total 96 ways.. This looks easy and also takes almost same amount of time..
Originally Posted by crab365
QIn how many ways can we put 7 different balls into 5 different boxes, such that no box is empty?
Why won't the following approach work?
First, we select 5 balls out of 7 in 7C5 = 21 ways.
These 5 balls are distributed in 5 boxes in 5! = 120 ways, now that 5 balls have been put in 5 boxes, one in each, so that no box remains empty.
The remaining 2 balls are put in the 5 boxes in 5X5 = 25 ways.
=>Ans.= 21X120X25 = 63000.
OA is given as 16800.
Please tell me where my approach is wrong.
You are repeating many cases.
Suppose balls are B1, B2, B3, B4, B5, B6, B7
Select 5 balls: 1,2,3,4,5 and put in 5 boxes
Now select 2 remaining balls and put in last box, it becomes, 1,2,3,4,567
Now another case:
Select 5 balls: 1,2,3,4,7
Now select 2 remaining balls and put in last box, it becomes. 1,2,3,4, 567
Both cases are SAME, however you have counted them as different, that's why you're getting a much larger answer.
Hope I made the confusion clear. 😃
Q1. In a box there are : 5 red balls, 4 green balls, 3 yellow balls, 2 pink balls, 1 black ball. In how many ways, 3 balls may be selected from the box?
Q2. There are 5 green balls, 4 yellow balls, 3 red balls in a box. In how many ways 4 balls maybe selected from box?
Thanks. Please be elaborate with the solution, I'm trying to understand some concepts, P&C; not one of my strong points 😞
sorry yar i am not getting your approach..try it
1,1,1,1,3 = 7C3*4!*5 = 4200
1,1,1,2,2 = 7C2*5C2*3!*10 = 12600
total = 16800
your approach is correct:)
See.. It is like this..
When you have 4 boys and 6 girls, you can pick:
Case 1: One boy and two girls or
Case 2: One girl and two boys
Now, when you pick one boy, one girl; you are left with 8 people. And you have to chose one of them. Say you choose a boy.
Now, first boy can be chosen in 4 ways. Say, B1 was chosen. And then from remaining 8, you chose B2.
Then there will be one case, when we choose B2 as first boy. And from remaining 8; we choose B1. So, we get same case.
So, like this, each case when we calculate like (4C1 * 6C1 * 8C1) will be counted twice.. Hence, divide by 2.
We can also solve it by making two cases:
Case 1: One boy and two girls => 4C1 * 6C2 = 4*15 = 60 ways
Case 2: One girl and two boys => 6C1 * 4C2 = 6*6 = 36 ways
Total 96 ways.. This looks easy and also takes almost same amount of time..
i thnk we cn also solve this problem in an easier way as follows:
total no. of ways of selecting 3 people=10c3 (containing all cases)
assumin we don't gt atleast 1 boy or 1 girl=(4c3+6c3)=24
so the total no. of cases with atleast 1 boy or 1 girl=10c3-(4c3+6c3)=120-24=96
no of solutions for 3x + y +z =30? where x,y,z>=0
How is this to be solved without binomial theorem?
Total number of ways 4 couples can be seated around circular table such that m
mbafrmfms SaysAnyone?????????????????
Total number of ways 4 couples can be seated around circular table such that men and women are seated alternately and no man seated beside his wife
ans: 1
2 8
3 7
4 6
5
suppose , position 1 will be occupied by a man,position 2 and 8 by a woman, then position one can be filled in 4 ways while 2 can be filled with 3c1 and 8 by 2c1 ways.Since it is a circle there is no start or end ,total no. of ways=(4 x 3c1 x 2c1 )/2
Seven different objects must be divded among three ppl. In how many ways can ths b done if 1 or 2 of them must get no objcts
a-381 b-36 c-84 d-174 e-180
ans a-381...plz post the expln
Seven different objects must be divded among three ppl. In how many ways can ths b done if 1 or 2 of them must get no objcts
a-381 b-36 c-84 d-174 e-180
ans a-381...plz post the expln
(i)1 person gets no objects.
So, 7 different objects need to be distributed in 2 ppl.
2 ppl can be chosen in 3C2=3 ways.
The objects can be distributed as follows:
(1,6)=7C6=7 ways.
(2,5)=7C5=21 ways.
(3,4)=7C4=35 ways.
(4,3)(5,2)(6,1) are the same, so total no of ways = 2X(7+21+35)=126.
Total cases for (i)=126X3=378(we multiply by 3 as there are total 3 choices for selecting the people).
(ii)2 ppl get no objects.
In this case, 1 person must get all objects. This person can be selected in 3C1=3 ways.
The objects can go to him in only 1 way.
So total cases for (ii)=3.
Answer (i)+(ii)= 378+3= 381.
(i)1 person gets no objects.
So, 7 different objects need to be distributed in 2 ppl.
2 ppl can be chosen in 3C2=3 ways.
The objects can be distributed as follows:
(1,6)=7C6=7 ways.
(2,5)=7C5=21 ways.
(3,4)=7C4=35 ways.
(4,3)(5,2)(6,1) are the same, so total no of ways = 2X(7+21+35)=126.
Total cases for (i)=126X3=378(we multiply by 3 as there are total 3 choices for selecting the people).
(ii)2 ppl get no objects.
In this case, 1 person must get all objects. This person can be selected in 3C1=3 ways.
The objects can go to him in only 1 way.
So total cases for (ii)=3.
Answer (i)+(ii)= 378+3= 381.
thanks...can u plz explain the concept of distributing different to similar and different to different with some conditions