Permutation and combination Exercise 3.4
1. DIVYA
2.SHANTHARAM
3. REPRESENTS
4. COMBINE
5. BALBHARTI
Answer-
2.You have two identical books on English, three identical books on Hindi and four identical books on mathematics. Find the number of distinct ways of arranging them on a shelf.
Answer-
3. Full stop a coin is tossed 8 times. In how many ways can we obtain
a)four heads and hotels
b) at least 6 heads?
Answer-
4. A bag has 5 red for blue and 4 green marbles. If all are drawn one by one and their colours are recorded how many different arrangements can be found?
Answer-
5. Find the number of ways of arranging letter of the word MATHEMATICAL how many e of these arrangements have all vowels together?
Answer-
6. Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have
a)letters R and h are never together
B) all vowels together?
Answer-
7. How many different words are formed if the letters are is used price and letter s and T are used twice each?
Answer-
8. find the number of arrangements of letter in the MUMBAI so that the letter B is always next to A.
Answer-
In the word MUMBAI the number of letters
is n 6 0f which M repeats twice, i.e., p = 2.
If B is always next to A, i.e., AB, then it forms
one unit and remaining 4 letters. Therefore
total number of letters n 5 and p = 2.
number of different ways of arrangements
of letters in which B is always next to A
`=\frac{\text{n!}}{\text{p!}}`
`=\frac{\text{5!}}{\text{2!}}`
`=\frac{\text{120}}{\text{2}}`
`=60`
9. Find the number of arrangements of letters in the word CONSTITUTION diet begin and end with an.
Answer-
There are 12 letters in the word
CONSTITUTION, in which '0', 'N', T repeat
two times each, 'T repeats 3 times.
When the arrangement starts and ends with
'N', other 10 letters can be arranged
between two N, in which '0' and T repeat
twice each and T repeats 3 times.
Therefore,Total number of arrangements with the letter N at the beginning and at the end `=\frac{\text{10!}}{\text{2!2!3!}}`
10. Find the number of different ways of arranging letters in the word arrange. How many of these arrangement do not have the two hours and to age together ?
In the word ARRANGE the number of letters
is n 7 of which A repeats twice, i.e., p = 2,
R repeats twice, i.e., q = 2 and rest are
distinct.
the number of ways in which the letters of
`=\frac{\text{n!}}{\text{p!q!}}`
`=\frac{\text{7!}}{\text{2!2!}}`
`=\frac{\text{5040}}{\text{4}}=1260`
Neither two R's nor two A's are together:
Let two R's and two A's occur together.
Considering two R's as one unit and two A'
as another unit we have 3+2 = 5 letters.
number of arrangements of letters in
number of arrangements of letters in
which two R's and two A's are together =
`=\frac{\text{5!}}{\text{2!2!}}`
`=\frac{\text{5x 4x 3x 2x 1}}{\text{2x 1x 2 x 1}}=39`
number of arrangements in which neither
two R's nor two As are together
= 1260-30
= 1230.
Permutation and combination Exercise 3.4
11.How many distinct 5 digit number can be formed using the digits 3, 2, 3, 4, 5.
Answer-
5 digit numbers are to be formed from 2,3,
2,3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4,5
OR 2, 3, 3, 4, 5
5!
Number of such numbers
`=\frac{\text{5!}}{\text{2!}}×2=5!=120`
Case II: Numbers are formed from 2, 2, 3,3
and any one of 4 or 5
Number of such numbers
`=\frac{\text{5!}}{\text{2!2!}}=60`
Required number =120+60
=180
distinct 5 digit numbers can be formed using the digit 3,2,3,2,4,5
12. Find the number of distinct numbers formed using the digits 3, 4 ,5 ,6 ,7, 8, 9 so that all positions are occupied by odd digits
Answer-
Odd places are 1st, 3rd, 5th and 7th and odd
digits are 3,5,7,9. Even places are 2nd, 4th
and 6th and even digits are 2, 4, 6
The four odd digits occupy 4 odd places in
P4 ways and three even digits can occupy
the remaining 3 even places in sP3 ways.
the total number of distinct numbers that
can be formed
`= ^{4}P_4×^{3}P_3`
`=4!×3!`
`= 4× 3×2 ×1 ×3 ×2×1`
`=144`
13. How many different 6 digit numbers can be formed using digits in the number 6, 5, 9 ,9 ,4 , 2? How many of them are divisible by 4?
Answer-
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Total number of arrangements
`=\frac{\text{6!}}{\text{2!}}`
`=\frac{6×5×4×3×2}{2}=360`
∴ 360 different 6-digit numbers can be formed.
For a number to be divisible by 4, the last two digits should be divisible by 4 i.e. 24, 52, 56, 64, 92, or 96.
Case I: When the last two digits are 24, 52, 56 or 64.
As the digit 9 repeats twice in the remaining four numbers, the number of arrangements
4!2!
= 12
∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48.
Case II: When the last two digits are 92 or 96.
As each of the remaining four numbers is distinct, the number of arrangements = 4! = 24
∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48.
∴ Total number of arrangements from both these cases is 48 + 48 = 96.
Thus, 96 6-digit numbers can be formed that are divisible by 4.
14. Find the number of distinct words formed from the letters in the word INDIAN. How many of them have the two ends together?
Answer-
In the word 'INDIAN', the number of letters is n = 6 of which I repeats twice, i.e., p = 2, N repeats twice, i.e., q = 2, and the rest are distinct.
∴ the number of different words formed with letters of the word INDIAN
`=\frac{\text{n!}}{\text{p!q!}}`
`=\frac{\text{6!}}{\text{2!2!}}`
`=\frac{\text{6×5×4×3×2!}}{\text{2×1×2!}}`
`=180`
Two N's are together:
Two N's form 1 unit and with the remaining 4 letters we have 5 letters in which I repeat twice.
Two N's form 1 unit and with the remaining 4 letters we have 5 letters in which I repeat twice.
∴ the number of different words formed in which two N's are together
`=\frac{\text{5!}}{\text{2!}}`
`=\frac{\text{5×4×3×2!}}{\text{2!}}=60`
15. Find the number of different ways of arranging letters in the word PLATOON if
a) if two O's are never together
b) if consonants and vowels occupy alternate
position.
Answer-
A) When the two O’s are never together
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
Two O’s can take their places in 6P2 ways.
But ‘O’ repeats 2 times.
∴ Two O’s can be arranged in
`=\frac{\text{6P2}}{\text{2!}}`
`=\frac{\text{6!(6-2)!}}{\text{2!}}`
`=\frac{\text{6×5×4!}}{\text{4!×2×1}}`
`=3×5`
`= 15 `ways
∴ Total number of arrangements if the two O’s are never together = 120 × 15 = 1800
B)In the word 'PLATOON', the number of letters is n = 7 of which 'O' repeats twice, i.e., p = 2
∴ the total number of words formed by using the letters of the word 'PLATOON'
`=\frac{\text{n!}}{\text{p!}}`
`=\frac{\text{7!}}{\text{2!}}`
`=\frac{\text{7×6×5×4×3×2!}}{\text{2!}}`
`= 2520`
Consonants and Vowels occupy alternate positions: There are 4 consonants P, L, T, N, and three vowels A, O, O.
The possible arrangement in which consonants and vowels take alternate places is CVCVCVC.
Three vowels can be arranged at three places in
ways (∵ O repeats twice) and four consonants can be arranged at 4 places in 4! ways.
∴ the number of such arrangements
`=\frac{\text{3!2!}}{\text{4!}}`
`=\frac{\text{3!2!}}{\text{4!}}{\text{4×3×2×1}}`
`=72`
Permutation and combination Exercise 3.4 Inlogicalmath.
