life of a woman after child birth

life of a woman after child birth

Constipation in the post partum stage usually affects some women (Some of the drugs used during the labour stage of pregnancy and the high levels of some hormones can contribute to the constipation ). So they might be prescribed stool softeners. The first few post partum poops after child birth might be terrifying and might cause anxiety for some new mothers making them very nervous and might be a traumatic experience for some of the new mothers . Sometimes the constipation starts after the woman returns to her house from the hospital.In extreme cases, manual disimpaction ( fragmentation ) using a gloved finger might be needed.   Fiber rich foods are good for dealing with constipation. Food rich in water, warm milk , oatmeal etc seem to be good for dealing with post partum constipation.

Painful hemorrhoids ,anal fissures after c-section  or vaginal birth also affect a lot of women in the post partum stage (feels very painful like glass / sharp knife passing through the anus and might be mixed with bleeding ) . Stool softeners and food rich in fiber, drinking lots of water seems to be helpful in dealing with hemorrhoids after c-section  or vaginal birth. Some of the affected women advice that even if stool does not come out when sitting on the loo, it is better not to strain too much for the bowel movement , because it might push the hemorrhoid out.

Some women suffer from vaginal dryness especially after childbirth due to the huge changes in hormone levels ( especially low levels of estrogen ) in the new mother ( the external vulva region can also get irritated for some women. ) resulting in burning / itching symptoms. Using a lubricant ( lube ) / moisturizer is usually recommended for such women.

A new baby can change a lot of things in your life. Some days after the child is born, the new mother will find that she has a very hectic life and she would usually like to have somebody with her to help her take care of the baby for at least some part of the day.  In the olden days her own mother or mother in law would be near her to share the responsibility of taking care of the new born child. But nowadays with more and more nuclear families , this type of support is also starting to lessen. That is when the importance of socializing with other stay at home mothers ( SAHM ) near your home becomes more important.

Some mothers feel a lot of anxiety in the days/weeks after child birth about coping with taking care of the child and her household. This is usually referred to as PPA ( post partum anxiety ) and may sometimes be mixed up with PPD ( post partum depression or post natal depression ).  Some of  these might be due to hormonal level changes in the post partum period.

Some parents just leave their baby with the grandparents of the child and go for weekend vacations / trips etc. After that the mother might wallow in self guilt, thinking  if it might not have been a better idea taking their child with them for the trip in terms of bonding with the child,  especially if the mother was breastfeeding the child . Some parents have the excuse that very small children will not remember the experience of the trip.

Then there is the question of when the new mother has to return to her work especially if she was a working woman before becoming pregnant.

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determine the number of 5 card selections of a deck of 52 cards if each selection of 5 cards should have exactly one king

ncert cbse chapter 7 permutations miscellaneous exercise

8.determine the number of 5 card selections of a deck of 52 cards if each selection of 5 cards should have exactly one king

First select the king. Since there are four kings, this can be done in

C(4,1) ways.

Now after removing the 4 kings there are 52-4 = 48 cards which will be used to select the remaining 4 non king cards which can be done in C(48,4)

Answer = C(4,1) * C(48,4)

9. It is required to seat 5 men and 4 ladies in a row so that the ladies occupy the even positions. How many such arrangements are possible.

The required arrangement is like

M W M W M W M W M

The men can be arranged in ( 5! ) =120 ways

Now arrange the ladies in their even positions in (4!) = 24 ways

Answer = 120*24 =2880 ways

10. In a class of  25 students, 10 are to be chosen  for an excursion party. There are 3 students who decide that either all of them will join or none of them will go. In how many ways can the party be chosen.

First possibility

All 3 of the friends join the party

Now all we have to do is to select the remaining 7 participants using the remaining (25-3)=22 students in C(22,7) ways

OR

Second possibility

None of the three friends join the party

Now we have to select the entire 10 participants using the remaining (25-3)=22 students in C(22,10) ways

Answer  = C(22,7) + C(22,10)

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

solution

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

solution

8.determine the number of 5 card selections of a deck of 52 cards if each selection of 5 cards should have exactly one king

solution

9. It is required to seat 5 men and 4 ladies in a row so that the ladies occupy the even positions. How many such arrangements are possible.

solution

10. In a class of  25 students, 10 are to be chosen  for an excursion party. There are 3 students who decide that either all of them will join or none of them will go. In how many ways can the party be chosen.

solution

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The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

ncert cbse chapter 7 permutations miscellaneous exercise

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

Because there are 5 vowels,

the 2 vowels can be selected in C(5,2) = [(5*4)/(1*2)] = 10 ways

Because there are 21 consonants

the 2 consonants can be selected in C(21,2) = [(21*20)/(1*2)] = 210 ways

After selecting the 4 required letters now we can arrange them in

(4!) = 4*3*2*1 = 24 ways

Required number of words = 10*210*24 =50400

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

let the selections be denoted by (x,y) where x denotes the questions from part I

and y the questions from part II

atleast 3 means 3 or more

(3,5) or (4,4) or (5,3) are the different possibilities with their respective number of

possible selections being

C(5,3) * C(7,5) = [(5*4)/(1*2)]*[(7*6)/(1*2)]=210 ways

C(5,4) * C(7,4) = [5]*[(7*6*5)/(1*2*3)]=175 ways

C(5,5) * C(7,3) = [1]*[(7*6*5)/(1*2*3)]=35 ways

Total number of ways = 210+175+35 =420 ways

Here note that C[n,r] =C[n,(n-r)]

gives

C[5,3]=C[5,2]

C[7,5]=C[7,2]

C[5.4]=C[5,1]=5

C[7,4]=C[7,3]

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

6.The english alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed

solution

7. In an examination  a question paper consists of two parts containing 5 and 7 questions respectively. A  student is required to answer 8 questions in all selecting atleast 3  from each part.  In how many ways can a student select the questions?

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

ncert cbse chapter7permutations miscellaneous exercise

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

first arrange the 11 letters in alphabetical order

AA   2 A's

E

II  2 I's

M

NN 2 N's

O

T

X

Therefore before the first word starting with E, words starting with the letter

A will be arranged in alphabetical order in the dictionary.

fix one A in the first slot

remaining 10 letters contain 2 I's and 2 N's [ one of the 2 A's is used for the first slot ]

So number of words starting with the letter A

= [(10!)/ {(2!)*(2!)}] =907200

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

The one's place should contain 0 for the number to be divisible by 10

Fix the digit 0 in the one's place Therefore the one's place can be filled in only one way.

The other 5 slots can be filled in 5! = 120 ways

Answer = 1*120 = 120

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

6.How many 6 digit numbers can be formed using 0,1,3,5,7,9 which are divisible by 10 and no digit is repeated?

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

ncert cbse 11th maths excercise 7.3 chapter 7

ncert cbse 11th maths  excercise 7..3 chapter 7

exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

There are 12 letters in the word PERMUTATIONS

i)words start with P and end with S

If the word start with P and end with S

first slot and 12th slot can be filled in 1 way each.

After removing P and S,

For the remaining 10 slots there are 10 items

with  2 T's,

So number of words starting with P and end with S

= 1 *[(10!)/(2!)]*1 = 1814400

ii)vowels are all together

vowels in the given word are E,U,A,I,O

If the vowels are together, think of the 5 vowels as a single block.

Inside the block the vowels can be arranged in (5!) ways

Now this block of vowels with the remaining 7 consonants

make up 7+1=8 pieces with 2 T's

which can be arranged in [(8!)/(2!)] ways

Number of words in which the vowels are together = (5!)*[(8!)/(2!)]

=120*20160=2419200

iii)there are always 4 letters between P and S

When there are 4 letters between P and S

one particular arrangement will be

P x x x x S # # # #

in this example first and fifth slots can be filled in 1 way each

and the remaining 10 slots can be filled with 10 remaining letters

containing 2 T's in = 1 *[(10!)/(2!)]*1 = 1814400 ways

Now possible re-arrangements  are

1        P x x x x S # # # ###

2        #P x x x x S # # # ##

3         ##P x x x x S # # ##

4        ###P x x x x S # ##

5        ####P x x x x S##

6        #####P x x x x S#

7        ######P x x x x S

Now the position of P and S can be interchanged

8       S x x x x P # # # ###

9        #S x x x x P # # # ##

10         ##S x x x x P # # ##

11        ###S x x x x P # ##

12        ####S x x x x P##

13        #####S x x x x P#

14        ######Sx x x x P

A total of 14 arrangements

number of words where

there are always 4 letters between P and S

=1814400*14

=25401600

 

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

ncert cbse 7th chapter permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

The vowels are A,U,E

We have to select 2 out of the 3 vowels in C(3,2)= [3*2]/[1*2]= 3 ways

The consonants are D,G,H,T,R

We have to select 3 out of the 5 vowels in C(5,3)= [5*4] / [1*2] =10ways

because C(n,r) = C[n,(n-r)]  gives C(5,3)=C(5,2)

now the 2+3= 5 selected letters can be arranged among themselves in

5! =5*4*3*2*1=120 ways

Required number of words = 3*10*120 =3600

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

Vowels are E,U,A,I,O

Treat the 5 vowels as a single unit. Inside this the 5 vowels can be arranged

among themselves in 5!=5*4*3*2*1=120 ways

Consonants are Q,T,N

Treat the 3 consonants as a single unit. Inside this the 3 consonants can be arranged among themselves in 3!=3*2*1=6 ways

Now the two units can be arranged among themselves in 2!=2*1 = 2 ways

Required number of words =120*6*2=1440 words

ncert cbse 11th mathematics chapter  7 permutations exercise 7.3

11.In how many ways can the letters of the word PERMUTATIONS be arranged if

i)words start with P and end with S

ii)vowels are all together

iii)there are always 4 letters between P and S

solution

ncert cbse 11th mathematics chapter  7 permutations miscellaneous exercise

1.How many words each of 2 vowels and 3 consonants can be formed using the letters of the word DAUGHTER

solution

2. How many words of can be formed using all the letters of the word

 EQUATION  so that the vowels and consonants occur together?

solution

4. If the different permutations of the word EXAMINATION are arranged in a dictionary how many words are there in this list before the first word starting with E

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookies

find the derivative of {ax+b}/{cx+d}

ncert cbse 11th limits and derivatives miscellaneous exercise

5.find the derivative of {ax+b}/{cx+d}

using quotient rule of differentiation

if y = (u/v)

dy/dx = {vu' - uv'} / {v^2}

u = {ax+b}; v = {cx+d}

u ' = a ; v ' = b

dy/dx ={(cx+d)(a) -(ax+b)(c)} / {(cx+d)^2}

dy/dx = {ad-bc} / {(cx+d)^2}

2. find the derivative of {px+q}{(r/x)+s}

using product rule of differentiation

 

if y =uv

dy/dx = uv' +vu'

u =px+q ; v = {(r/x)+s}

u' = p(1)+0 = p  ; v ' = {{-r} / {x^2} + 0] = {-r} / {x^2}

y = {px+q}{(r/x)+s}

dy/dx = uv' +vu'

dy/dx = {px+q}{{-r} / {x^2}} +{(r/x)+s}{p}

dy/dx = {{-pr}/ {x} } -{{qr} / {x^2}} +{{rp}/x} +sp

dy/dx = -{{qr} / {x^2}} +ps

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution

2. find the derivative of {px+q}{(r/x)+s}

solution

5.find the derivative of {ax+b}/{cx+d}

solution

7. find the derivative of {1} / {a x^2 + bx + c }

solution

8.find the derivative of {ax+b}/{p x^2+ qx +r}

solution

17.find the derivative of (sinx+cosx) / (sinx-cosx)

solution

18. find the derivative of (secx -1) / (secx+1)

solution

23.find the derivative of [(x^2)+1][cosx]

solution

24. find the derivative  of (a(x^2)+sinx )[p+qcosx]

solution

28.

find the derivative of x / (1+tanx)

solution

29.

find the derivative of (x+secx)(x-tanx)

solution

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There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks  using tracking cookie