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

(i) Words start with P and end with S,

(ii) Vowels are all together,

(iii) Thereare always 4 letters between P and S?

22. There are 12 letters in which T appears 2 times and rest are all different.

i. When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.

Number of permutation = $\frac{10!}{2!}$

= 18,14,400

ii. We take the 5 vowels (E, U, A, I, O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T's.

So, number of permutations in which the vowels come together

= permutation of 8 object x permutation within the vowels

= $\frac{8!}{2!}$ * 5!

= 20160 * 120

= 2419200

iii. In order to have 4 letters between P and S, (P, S) should have the possible sets of places (

...more

22. There are 12 letters in which T appears 2 times and rest are all different.

i. When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.

Number of permutation = $\frac{10!}{2!}$

= 18,14,400

ii. We take the 5 vowels (E, U, A, I, O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T's.

So, number of permutations in which the vowels come together

= permutation of 8 object x permutation within the vowels

= $\frac{8!}{2!}$ * 5!

= 20160 * 120

= 2419200

iii. In order to have 4 letters between P and S, (P, S) should have the possible sets of places (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11) and (7, 12) similarly (S, P) should have the same possible sets of places.

So, total number of possible positions fixed for P and S = 14

After fixing P and S we can rearrange the remaining 10 places with the remaining 10 letters of which 2 are T's.

So, required permutation = $\frac{10!}{2!}$

= 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3

= 18,14,400

Therefore, total permutation for which there are always 4 letters between P and S

= 14 * 18,14,400

= 2,54,01,600

less

<p><strong>22. </strong>There are 12 letters in which T appears 2 times and rest are all different.</p><p><strong>i. </strong>When P and S are fixed as first and last letter we can arrange the remaining 10 letter taking all at a time. i.e.</p><p>Number of permutation =&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>1</mn><mn>0</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo></mrow></mfrac></mrow></math></span></p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1747817863phpJdKUMu.jpeg" alt="" width="290" height="66" loading="lazy"></picture></div></div><p>= 18,14,400</p><p><strong>ii. </strong>We take the 5 vowels (E, U, A, I, O) as one single object. This single object with the remaining 7 object are treated as 8 object which have 2 – T's.</p><p>So, number of permutations in which the vowels come together</p><p>= permutation of 8 object x permutation within the vowels</p><p>=&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>8</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo></mrow></mfrac></mrow></math></span>&nbsp;* 5!</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1747817950phpDkRyOF.jpeg" alt="" width="317" height="51" loading="lazy"></picture></div></div><p>= 20160 * 120</p><p>= 2419200</p><p><strong>iii. </strong>In order to have 4 letters between P and S, (P, S) should have the possible sets of places (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11) and (7, 12) similarly (S, P) should have the same possible sets of places.</p><p>So, total number of possible positions fixed for P and S = 14</p><p>After fixing P and S we can rearrange the remaining 10 places with the remaining 10 letters of which 2 are T's.</p><p>So, required permutation =&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>1</mn><mn>0</mn><mo>!</mo></mrow><mrow><mn>2</mn><mo>!</mo></mrow></mfrac></mrow></math></span></p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1747818032php12BKV9.jpeg" alt="" width="261" height="47" loading="lazy"></picture></div></div><p>= 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3</p><p>= 18,14,400</p><p>Therefore, total permutation for which there are always 4 letters between P and S</p><p>= 14 * 18,14,400</p><p>= 2,54,01,600</p>

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