MadeEasy Full Length Test 2019: Combinatory - Permutations And Combinations

Question

The number of ways 5 letter be put in 3 letter boxes A,B,C. If letter box A must contain at least 2 letters.

Comments

I got the same but its wrong :/

---

270 ???

---

No its 131

---

yes its 131..?

---

kaise ?

---

Please Discuss the Approach

---

Why cant i write :-X1 + X2 + X3 = 5 and X1 >= 2

Assuming we have identical letters and different letter boxes

---

@Na462 posted :) letters are different always unless they specify

---

is answer 10?

---

no, it is 131

---

madeeasy's answer might be wrong

---

x1+x2+x3=5

given condition is that one box contain at least 2 letters which is x1>=2

x1-2>=2-2

x1-2>=0

we can write 

x1-2+x2+x3=5

x1+x2+x3=7

using the concept of stars and bars we can determine 

||+||+|||=7

9c2=36

what is wrong in this ?

@Shaik Masthan @MiNiPanda

---

@vijju532

It specifically mentioned box A has atleast 2 letters. Its not right to consider "one box".

Let the 5 letters be named as L1,L2,L3,L4,L5.

Case 1) Put 2 letters in A. Ways of choosing 2 letters from 5 is C(5,2).

No. of ways of putting remaining 3 letters into B and C is $2^3$.

Let A has L1 and L2. So remaining is L3 to L5.

L3 can go to B or C. 2 ways

L4 can go to B or C. 2 ways.

L5 can go to B or C. 2 ways.

So total cases here is $C(5,2)x2^3 =10*8=80$

Case 2)

Put 3 letters in A. Ways of choosing 3 letters from 5 is C(5,3).

No. of ways of putting remaining 2 letters into B and C is $2^2$.

So total cases here is $C(5,3)x2^2 =10*4=40$

Case 3)

Put 4 letters in A. Ways of choosing 4 letters from 5 is C(5,4).

No. of ways of putting remaining 1 letters into B and C is $2^1$.

So total cases here is $C(5,4)x2^1 =5*2=10$

Case 4)

Put 5 letters in A. Ways of choosing 5 letters from 5 is C(5,5).

No. of ways of putting remaining 5 letters into B and C is $2^0$.

So total cases here is $C(5,5)x2^0 =1*1=1$

Sum them up: 80+40+10+1=131

 

---

can we use the concept of onto funtion ?

---

@vijju532 I edited the comment. You can check it. Yes concept of on-to can be used.

---

Another method

$3^{5}$ - ( {when box 1 contain only 1 letter } - {when box 1 contains 0 letter} )

---

@Magma Can u please explain in more details. I am getting $3^5 - 2^4 - 2^5 = 195$

Answer 1

 5 letters be put into 3 letter boxes A,B,C...

// means 5 different letters in 3 different boxes..

condition : letter box A contains at least 2 letters...

method 1:

required ways where letter box A has at least 2 letters= box A has 2 letters + box A has #letters + box A has 4 letters + box A has 5 letters

                         =5C2*2^3 + 5C3*2^2+5C4*2^1 +5C5

                        =80+40+10+1 =131 answer...

shortcut method:

required ways where letter box A has atleast 2 letters=total ways - at most one letter in box A

= total number of ways in which each letter can be assigned to any letter box - when letter box A has 1 letter - letter box A has no letter.

=> 3^5 - 5C1*2^4 - 5C0*2^5

=243- 5*16 - 32

=243-112 = 131 answer.... 

 

Comments

@Magma kuch time pehle bhot disturb kie hai aise questions to never let the same mosquito bite you twice... :p

@Na462 thanks bro :)

---

@arvin waah bhai....lgta hai iss baar tum hi top kroge!

---

@himgta ni bro revision ni hopara dhang se aur test series v ni die acche se :/ work load bhot hai :/ 

Answer 2

The answer depends on weather the letters are same or different. If the letters are same answer is 10, if different answer is 131.

Comments

According to que 131 is the right answer.

---

You mean $10$ is possible as well?

---

Yes it is 10 if the letters are indistinguishable.

Answer 3

correct answer is 10.

This problem belongs to combinations with repetitions. Try the examples from rosen's book