Permutations & Combinations - Questions & Discussions

MBA CAT 2022
672
@chillfactor said:
Question says that 90 is qualifying marks and in how many ways one can qualify, so he can get more than 90 also.
thats why greater than equal to sign is there.
Haan but when a + b+c +d = 90
Tab (n+r-1)C(r-1) ways hote hain na Sir toh 93C3 hoga na ???

Similarly 47C3 hoga na ? I guess so :/
Lotcha ho toh bata do.
673

@Estallar12 said:
Haan but when a + b+c +d = 90 Tab (n+r-1)C(r-1) ways hote hain na Sir toh 93C3 hoga na ???Similarly 47C3 hoga na ? I guess so :/ Lotcha ho toh bata do.
@[571257:Estallar12] + b+c +d ‰¤ 90 (less than equal to)
so, a + b + c + d + e = 90 (dummy variable)

thats why its C(94, 4) ways and same way for the unaccepted cases
674
@chillfactor said:
@Estallar12 + b+c +d ‰¤ 90 (less than equal to) so, a + b + c + d + e = 90 (dummy variable)
thats why its C(94, 4) ways and same way for the unaccepted cases
Phir toh a + b + c + d +e
Agar a + b+ c + d+e so, a + b + c + d + e = 90 (dummy variable) karenge toh a + b +c + d ka sum zero bhi ho sakta hai na when e = 90.
That's not acceptable na Sir ? :O
675

@Estallar12 said:
Phir toh a + b + c + d +e so, a + b + c + d + e = 90 (dummy variable) karenge toh a + b +c + d ka sum zero bhi ho sakta hai na when e = 90. That's not acceptable na Sir ?
marks are 45 - a, 45 - b, and so on
so, a can not be more than 45, and these cases are subtracted later on.
676
@neerajagayatri said:

There are 4 identical oranges, 3 identical mangoes and 2 identical apples in the basket. the number of ways in whihc we can select one or more fruits from the basket isAns:60, 59, 57, 55, 56

Please help in solving this

With oranges i have 5 chances namely -> picking one orange or two oranges or three oranges or four oranges and not picking up an orange at all i.e. 5 chances
Similarly with mangoes 4 chances and with apples 3 chances.

Total chances = 5X4X3 = 60. But I cannot leave out all of them since i need to pick at least one fruit.

Hence, the number of ways = 60-1 = 59
677
in how many ways can a delegation of 4 professors and 3 students be constituted
from 8 professors and 5 students if balamurli an arts student refuses to be in the
delegation when prof. siddharth the science professor is included in it
(A)280
(B)210
(C)490
(D)560
678

Find the no of ways of selecting n articles out of 3n+1 out of which n are identical...

679

Hi All,

7 different objects must be divided among 3 people . In how many ways can this be done if one or two of them can get no objects

For this ans is given as 2187.. nothing but 3^7... but i think that contains a combination where each person gets 0,0,0 objects and i feel that we need to subtract that from the total number of possibilities.

Please correct me if i am wrong...

Thanks in advance

680
@Swapzilla said: In an examination , max marks for 3 papers is 50 each and for the fourth is 100 . Find the number of ways in which a student can score 60% aggregate marks .

A. 330850
B. 233551
C. 110551
D. 220800


there are two possible solution


solution 1 :-


the question is same for find out the coefficient of x^150 in the expansion of the expression

(1 + x+x^2 + x^3 + x^4 + ......... x^50 )^3 (1 +x + x^2 +.................+ x^100)

= (1 - x^51)^3 * (1 - x^101) (1-x)^-4

now find out coefficient of x^150 in the given expression

= (1 - 3 x^51 + 3 x^102 - x^153) (1 - x^101) ( 1 + 4c1 x + 5c2 x^2 + ............... )
= (1 - 3 x^51 + 3 x^102 - x^101 + ......... ) ( 1 + 4c1 x + 5c2 x^2 + ............... )
now coefficient of x^150
= 153c150 - 3 * 102c99 + 3 * 51c48 - 52 c49
= 153 c 3 - 3 102 c 3 + 3 * 51 c3 - 52 c 3
= 110551

681
@ajeetaryans said:
there are two possible solutionsolution 1 :-the question is same for find out the coefficient of x^150 in the expansion of the expression(1 + x+x^2 + x^3 + x^4 + ......... x^50 )^3 (1 +x + x^2 +.................+ x^100)= (1 - x^51)^3 * (1 - x^101) (1-x)^-4now find out coefficient of x^150 in the given expression= (1 - 3 x^51 + 3 x^102 - x^153) (1 - x^101) ( 1 + 4c1 x + 5c2 x^2 + ............... ) = (1 - 3 x^51 + 3 x^102 - x^101 + ......... ) ( 1 + 4c1 x + 5c2 x^2 + ............... ) now coefficient of x^150 = 153c150 - 3 * 102c99 + 3 * 51c48 - 52 c49 = 153 c 3 - 3 102 c 3 + 3 * 51 c3 - 52 c 3 = 110551
ii solution

let he want to earn 150 marks so reduce a
100 marks in how many ways he can lose 100 marks out 250
(50 - x1) + ( 50 - x2 ) + (50 - x3 ) + (100 - x4) = 150
x1 + x2 + x3 + x4 = 100

103 c 3
and x1 , x2, x3 are less than or equal to 50 and x4 is less than 100

first calculate for x1
when x1 > 50
let x1 = 51 + x put in the question

x + x2 + x3 +x4 = 49
52 c 3

so same for x2 and x3

so ans

103c3 - 3 * 52 c 3

= 110551



682

please help with these questions

Six letters are to be placed in six addressed envelopes. If the letters are placed at random into the envelopes, the probability that
None of the six letters are placed into their corresponding envelopes.

Seven letters are to be placed in seven addressed envelopes. If the letters are placed at random into the envelopes, the probability that
Exactly three of the seven letters are placed into their corresponding envelopes.

683
@meenag3 said: please help with these questionsSix letters are to be placed in six addressed envelopes. If the letters are placed at random into the envelopes, the probability that None of the six letters are placed into their corresponding envelopes.
1) concept derangement : 6! ( 1/2! - 1/3! + 1/4 ! - 1/5! + 1/6! ) = 360 - 120 + 30 - 6 + 1 = 265 .

sample space = 6! = 720 .. so prob = 265 / 720

684
@naga25french said:
1) concept derangement : 6! ( 1/2! - 1/3! + 1/4 ! - 1/5! + 1/6! ) = 360 - 120 + 30 - 6 + 1 = 265 .sample space = 6! = 720 .. so prob = 265 / 720

thnk you for the reply...however i have few doubts on the usage of this formula...this is used gnerically for those cases where all envelopes are misplaced ,is it?
685

How many arrangements of six 0's, five 1's, and four 2's are there in which the first 0 precedes the first 1, precedes the first 2? Pl help.

686

@[495335:meenag3] is the answer 12!..

687
@meenag3 said:

How many arrangements of six 0's, five 1's, and four 2's are there in which the first 0 precedes the first 1, precedes the first 2? Pl help.

required arrangement is 

5 0's, 4 1's, 4 2's
4 0's, 4 1's, 4 2's
3 0's, 4 1's, 4 2's
2 0's, 4 1's, 4 2's
1 0's, 4 1's, 4 2's
0 0's, 4 1's, 4 2's

counting for all six cases : (13!/ 5!4!4! + 12!/4!4!4! + 11!/3!4!4! + 10!/2!4!4! + 9!/4!4! + 8/4!4! )

= 140140










688
@meenag3 said: please help with these questionsSix letters are to be placed in six addressed envelopes. If the letters are placed at random into the envelopes, the probability that None of the six letters are placed into their corresponding envelopes.Seven letters are to be placed in seven addressed envelopes. If the letters are placed at random into the envelopes, the probability that Exactly three of the seven letters are placed into their corresponding envelopes.
Q.1 None of the six letters are placed into their corresponding envelopes
solution:-

this is the concept of inclusion and excursion theory (set theory )
6! - 6c1*5! + 6c2 * 4! - 6c3 * 3! + 6c4 * 2! - 6c5 * 1! + 6c6 0!

= 720 - 720 + (15 * 24 ) - (20 * 6) + (15 *2) - 6 +1

= 0 + 360 -120 +30 -6

= 264

Seven letters are to be placed in seven addressed envelopes. If the letters are placed at random into the envelopes, the probability that
Exactly three of the seven letters are placed into their corresponding envelopes

= 7c3 *4! - 7c4 * 3! + 7c5 *2! - 7c6 * 1! + 7c7 *0!



689

four identical bags are distributed among four boys.If each boy can get any no of bags,find the probability no boy gets more than two bags??

1. 18/35 2. 2/7 3.19/35 4. 16/35

690
@NishantGupta23 said: four identical bags are distributed among four boys.If each boy can get any no of bags,find the probability no boy gets more than two bags??1. 18/35 2. 2/7 3.19/35 4. 16/35
Possible cases ...

4!/3!=4
4!/2!=12
4!/2!=12
4!/=6
4!/4!=1

total=35..

probability=19/35...
691
hi !!

Pls. help with the detailed sol for the below mentioned probs of POM COM:-


Q1) the crew of an 8 member rowing team is to be chosen from 12 men, of which 3 must row on 1 side and 2 must row on the other side only. find the number of ways of arranging the crew with 4 members on each side ??


Options :- A) 40320 B) 30240 C) 60480 D) 10080 E) None of these


Q2) in how many ways 5 MBAs and 6 CA students can be arranged together so that no 2 MBA students are side by side:-


Options:-a) 7!*6!/2! b) 6!*6! c) 6!*5! d) 11P5 e) 11C5


Q3) there are 8 orators A,B,C,D,E,F,G and H. In how many ways can the arrangements be made so that A comes before B and B comes before C alwaz??


Options: -A) 8!/3! B) 8!/6! C) 5!*3! D) 6!*3! E) 8!/(5!*3!) F) None of these


Q4) if we have to make 7 boys and 7 girls sit around a round table which is numbered, then the number of ways in which it can be done are??


Options - A) 2*7!*7! B) 7!*6! C) 7!*7! D) None of these


Q5) in a building there are 12 floors excluding the ground floor. 9 ppl get in the lift at the ground floor. the lift does not stop at the 1st floor. 2,3,4 ppl alight from the lift on its upward journey, then how many ways then can do so?


Options- A) 11C3* 3! B) 11P3*9C4*5C3 C) 10P3*9C4*5C3 D) 12C3


Q6) 7 different objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??


Options - A) 15120 B) 2187 C) 3003 D) 792