# TCS Permutations and Combinations Questions with Solutions – 1

1.In how many ways a team of 11 must be selected from 5  men and 11 women such that the team must comprise of not more than 3 men?
a. 1565
b. 1243
c. 2256
d. 2456
Answer: C
Explanation;
The team may consist of 0 men + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women.
So Number of ways are = 11C11+5C1×11C10+5C2×11C9+11C11+5C1×11C10+5C2×11C9+5C3×11C85C3×11C8 = 2256

2.The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the two sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged.
Answer: 2×7×4!×14C2×4!
Explaination:We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7×4! ways.
Now for the first side we need two people from the remaining 14. So this can be done in
14C2 ways and this side people can sit in 4C2 ×4! ways.
Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4!

3.How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?
a.30
b.35
c.55
d.60
Answer:60
Explaination: Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6
There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways
so total number of ways of arranging all these numbers are 10 * 6 = 60 ways

4.The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?
a.AOTDSP
b.AOTPDS
c.AOTDPS
d.AOSTPD
Answer:d
Explaination:In alphabetical order : A D O P S T
A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A.
A D _ _ _ _ : empty places can be filled in 4!=24
A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
A O D _ _ _ : 3!= 6
A O P _ _ _ : 3!=6
Till this 36 words are obtained, we need the 42nd word.
AOS _ _ _ : 3!= 6
Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
So given word is AOSTPD

5.Total number of 4 digit number do not having the digit 3 or 6.
a.3342
b.3584
c.3241
d.3458
Answer:b
Explaination: consider 4 digits _ _ _ _
1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0)
2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9)
3rd blank can be filled in 8C1 ways
4st blank can be filled in 8C1 ways
Therefore total 4 digit number without 3 or 6 is 7 x 8 x 8 x 8=3584

6.What is the 32nd word of “WAITING” in a dictionary?
a.AGNTIWI
b.AGNWITI
c.AWGNTIW
d.ANGWITI
Answer:a
Explaination: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W
Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways
so can’t be arranged starting with A alone as it is asking for 32nd word so it is out of range
AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words.
AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24
AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways
so,24+12 =36th word so out of range. So we should not consider all the words start with AGN
now AGNI_ _ _can be arranged in 3! ways =6 ways
so 24+6=30 within range
Now only two word left so, arrange in alphabetical order.
AGNTIIW – 31st word
AGNTIWI – 32nd word

7.Find the total number of combinations of 5 letters a,b,a,b,b taking some or all at a time?
a.12
b.11
c.10
d.9
Answer: b
Explaination:1 letter can be chosen in 2 ways. a or b
2 letters can be chosen in 3 way. aa, ab, bb
3 letters can be chosen in 3 ways. bbb, aab, bba
4 letters can be chosen in 2 ways. aabb, bbba
5 letters can be chosen in 1 way.
So total ways are 11

8.How many positive integers not more than 4300 of digits 0, 1, 2, 3, 4 if repetition is allowed?
a.560
b.565
c.575
d.625
Answer:c
Explanation: one digit no =4 (0 is not a positive integer) two digit no=4*5=20 three digit no=4*5*5=100 four digit no=3*5*5*5=375(the possibility for 1,2,3 will come in the first position) four digit no=1*3*5*5(the possibility of 4is fixed in the first position and then 0,1,2 is comes in second position) and the last digit is 4300 we include this number also Ans is 4+20+100+375+75+1=575

9.In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.?
a.1234
b.1565
c.2456
d.2256
Answer:d
Explanation: Maximum 3 men can be played which means there can be 0, 1, 2, 3 men in the team. (5C0×11C11)+(5C1×11C10)+(5C2×11C9)+(5C3×11C8)=2256

10.Find the number of ways a batsman can score a double century only in terms of 4’s & 6’s?
a.15
b.16
c.17
d.18
Answer:b
Explanation: different ways 4x+6y=200 x=0 is false since 200 should be summation of both 4’s and 6’s. (x,y)=(47,2),(44,4)……..(2,32) so they are total 16 WAYS… 4’s and 6’s ———— 50 0 47 2 44 4 41 6 38 8 35 10 32 12 29 14 26 16 23 18 20 20 17 22 14 24 11 26 8 28 5 30 2 32 So total 17 ways but here it is 4’s & 6’s both so don’t consider 1st one Final ans : 16 ways