Permutation Combination-Cw1

PERMUTATION COMBINATION-CW1

1 INFOMATHS/MCA/MATHS/

NIMCET-2015

1.If 42 (nP2) = nP4 , then the value of n is

(a) 2(b) 4(c) 9(d) 42

PUNE-2015

2.The maximum number of cricket ball, which are placed, such that touches the remaining is 4. Then the maximum number of coins that are placed such that touches the remaining is

(a) less than 5

(b) greater than 4 but less than 8

(d) None

PU-2015

3.There are 10 true false questions. The number of ways in which they can be answered is :

(a) 10!(b) 210(c) 102(d) 100

4.The number of diagonals in an n sided figure is equal to :

(a) nC2 – 2 (b) nC2

5.The number of all possible words that can he formed using the letters of the word “MATHEMATICS” is :

(a) 11! (b) 11! / (2!2!2!)

NIMCET-2014

6.A password consists of two alphabets from English followed by three digits chosen from 0 to 3. Repetitions are allowed. The number of different passwords is

(a) 26P125P24P13P12P1

(b) (26P1)2 (4P1)3

(d) (26P14P1)2

7.The number of ways in which 5 days can be chosen from each of the 12 months of a non-leap year is

(a) (30C5)4 (31C5)7 (28C5) (b) (30C5)6 (31C5)6

8.The number of ways to arrange the letters of the English alphabet so that there are exactly 5 letters between the letters ‘a’ and ‘b’ is

(a) 24P5(b) 24P5 20!

9.If n and r are integers such that 1 ≤ r ≤ n then the value of n(n-1Cr-1) is

(a) nCr(b) r(nCr)

10.There are 8 students appearing in an examination of which 3 have to appear in Mathematics paper and the remaining 5 in different subjects. Then the number of ways they can be made to sit in a row if the candidates in Mathematics cannot sit next to each other is

(a) 2400(b) 16200(c) 4200(d) 14400

11.How many even integers between 4000 and 7000 have four different digits?

(a) 672(b) 840(c) 504(d) 728

MP-2014

12.In how many different ways can the letters of the word ‘ABILITY’ be arranged?

(a) 1260(b) 2520(c) 2420(d) 720

13. is

(a) – 1 (b) 2k(c) 2n(d) 0

PU-2014

14. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels canbe formed?

(A) 1050 (B) 210(C) 25200(D) 2520

15. In how many different ways can the letters of the word ‘COMPARE’ be arrangedsuch that the vowels always come together?

(A) 256 (B) 720(C) 360 (D) 480

16. From a group of 7 boys and 6 girls, five persons are to be selected to form a committeeso that at least 3 boys are there in the committee. In how many ways selection ispossible?

(A) 564 (B) 735(C) 756 (D) 645

17. In how many ways can the letters of the word ‘BEATLE’ be arranged?

(A) 720 (B) 120(C) 180 (D) 360

Bhu-2014

18.In an experiment, a coin is tossed twice. If the second toss results in a head, a die is rolled. The number of elements in the sample space is :

(a) 9(b) 12

19.The value of is :

(a) 2n–1(b) 2n

20.The sum of the digits in the unit place of all the four digit numbers formed with 2, 3, 4, 5 taken all at a time, is :

(a) 14(b) 42

21.There are 4 letters and 4 directed envelops. The number of ways all the letters are placed in a wrong envelope is :

(a) 6(b) 8

22.Four men and four women are to sit around a circular table such that there is a man on either side of every woman. The number of seating arrangements is :

(a) 3!  4! (b) (3!)2

NIMCET-2013

23.Let Tn denote the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If Tn+1 – Tn = 21 then n equals

(a) 5(b) 7(c) 6(d) 4

24.In how many different ways can the letters of the word “CORPORATION” be arranged so that all the vowels always come together?

(a) 810(b) 1440(c) 2880(d) 50400

25.The number of non-negative integers less than 1000 that contain the digit 1 are

(a) 92(b) 93

26.The total number of numbers that can be formed using the digits 3, 5 and 7 only if no repetitions are allowed, is

(a) 39 (b) 105(c) 15(d) 27

27.The sum of integers between 200 and 400, that are multiples of 7 is

(a) 8729 (b) 8700 (c) 8972 (d) 8279

PU-2013

28. How many license plates can be made using either 2 letters followed by 3 digits or 3 letters followed by 2 digits?

(A) 2433600 (B) 676000

29. In how many ways can 3 boys and 3 girls sit in a row if all the boys sit together, and all the girls sit together?

(A) 18 (B) 36(C) 72 (D) 90

BHU-2013

30.At an election, a vector may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is BHU-2013

(a) 5040(b) 6210(c) 385(d) 1110

PERMUTATION COMBINATION CW-1

1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10
C / A / B / C / B / B / A / C / B / A
11 / 12 / 13 / 14 / 15 / 16 / 17 / 18 / 19 / 20
B / - / - / C / B / C / D / - / - / -
21 / 22 / 23 / 24 / 25 / 26 / 27 / 28 / 29 / 30
- / - / B / D / D / C / A / A / C / -

SOLUTIONS

NIMCET-2015

1.Ans. (c) We have, 42 nP2 = nP4

 n2 – 5n + 6 = 42

 n2 – 5n – 36 = 0

 (n – 9) (n + 4) = 0

n = 9

PUNE-2015

2.ans. (a)

PU-2015

3.Ans. (b) 210

4.Ans. (c)nC2 – n

5.Ans. (b) as 2 times,

H 1

E 1

I  1

C1

S 1

NIMCET-2014

6.Ans. (b) ______

Ist position can be filled in 26P1 ways.

IInd position can be filled in 26P1 ways.

IIIrd position can be filled in 4P1 ways.

IVth position can be filled in 4P1 ways.

Vth position can be filled in 4P1 ways.

Hence, the password can be made.

(26P1)2 . (4P1)3 ways.

7.Ans. (a) In a non leap year.

There are 7 months with 31 days.

4 months with 30 days.

1 months with 28 days.

 no of ways = (30C5)4 . (31C5)7 . (28C5).

8.Ans. (c) 5 letters away 24 letters can be arranged in 24P5 ways  2

9.Ans. (b) n . n-1Cr-1 = r.nCr

10.Ans. (a) As no maths student must sit together

 Remaining 5 students of different subjects can be seated in 5! Ways & 3 maths students can be seated in remaining 6 positions in 6P3 ways.

 Total ways = 5! 6P3

= 120  6.5.4 = 120  120 = 14400

11.Ans. (b)

Case i) Left position is filled by 4.

Units place may be filled in 4 ways S{0, 2, 6, 8}

Tens place may be filled in 8 ways

Hundreds place may be filled in 7 ways

Total ways = 4  8  7 = 224

Case ii) Left position is filled by 5 units place filled in 5 ways {0, 2, 4, 6, 8}

Tens placed filled in 8 ways

Hundreds placed filled in 7 ways

Total ways = 5  8  7 = 280 ways.

Case in left position filled by 6 same as case i)

Total no. of ways 224 + 280 + 224 = 728.

MP-2014

12.

13.

PU-2014

14.Ans. (c)Out of 7 constants and 4 vowels. 3 constants and 2 vowels may be selected in 7C34C2 ways.

Further, they may be arranged in 5 ways to form distinct words.

 Total no. of words = 7C34C2 5! = 25200.

15.Ans. (b)Let the vowels be a single word.

i.e we treat OAE as a single letter

Now, there are 5 letters C, M, P, R, OAE which may be arranged in 5! Ways.

Also, OAE may be internaly arranged. in 3! Ways.

 Total no. of words = 5!  6 = 720 words.

16.Ans. (c)There are 7 boys and 6 girls, the chosen 5 persons may include 3 boys and 2 girls or 4 boys and 1 girl or 5 boys only.

 Total no. of ways of selection.

= 7C3.6C2 + 7C4.6C1 + 7C5

= 35.15 + 35.6 + 21 = 256 ways

17.Ans. (d)All letters of word BEATLE may be arranged in ways.

Bhu-2014

18.S.Space {TT, HT, HH1,TH1,

HH2, TH2,

HH3, TH3,

HH4, TH4,

HH5, TH5,

HH6, TH6}

Here, the sample space contains 14 elements

19.

= 2n – 1

20.Sum of digits of units place

= Sum of deef  (n – 1)!

= 14  3!

= 14  6

= 84

21.4 letters 5 can be placed in 4 envelops incorrectly in

= 9

22. 4 men can be seated in a circular table in 3! Ways.

Now, 4 women can be seated in 4 seats in 4! Ways

 Total no. of ways = 3!.4!

NIMCET-2013

23. Ans. (b)According to the question we have

n+1C3 – nC3 = 21

We know 8C3 – 7C3 = 21  n = 7

24. Ans. (d)Required ways

= 50400

25. Ans. (d)The numbers containing 1 is 103 - 93.

26. Ans. (c)Single digit numbers are 3.

Two digit numbers are 3×2 = 6

Three digit numbers are 3×2×1 = 6

Total numbers are 15.

27. Ans. (a)The integers will be203 + 210 + ……+ 399

There are terms in the series.

Sum

PU-2013

28.Ans. (a) Case (i) If the licence plates have 2 letters with 3 digits. Then the letters may be arranged in 262 ways and the digits may be chosen in 103 ways.

 No. of licence plates with 2 letters followed by 3 digits = 262 103

Case (ii) If licence plates have 3 letters with 2 digits

Then, the letters may be arranged in 263 ways and 2 digits may be arranged in 102 ways.

 No. of licence plates with 3 letters and 2 digits = 263 102

So, Total no. of licence plates

= 262.103 + 263.102 = 262.102(26+10)

= 262 36  100 = 2433600 licence plates.

29.Ans. (c)As there are 3 boys and 3 girls taking all boys together and considering them as a single unit. Similarly taking all girls as a single unit.

All boys together and all girls together can be seated in 2! 3! 3! Ways = 72 ways

BHU-2013

30.Ans. ()

1 INFOMATHS/MCA/MATHS/