21. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

21. There are 11 letters of which M appears 1 time, I appears 4 times, S appears 4 times and P appears 2 times.

The required number of arrangements = $\frac{11!}{4!4!2!}$

= 11 * 10 * 9 * 5

= 34650

When the four I occurs together we treat them as single object IIII. This single object together with 7 remaining object will account for 8 object which have 1-M. 2-P and 4-S.

So, required number of permutation = $\frac{8!}{4!2!}$$\frac{\mathrm{}}{}$

= 840

Therefore, total no. of permutation in which 4-I's do not come together

= 34650 – 840

= 33810

<p><strong>21. </strong>There are 11 letters of which M appears 1 time, I appears 4 times, S appears 4 times and P appears 2 times.</p><p>The required number of arrangements =&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>1</mn><mn>1</mn><mo>!</mo></mrow><mrow><mn>4</mn><mo>!</mo><mn>4</mn><mo>!</mo><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/1747812961php4h50rC.jpeg" alt="" width="235" height="57" loading="lazy"></picture></div></div><p>= 11 * 10 * 9 * 5</p><p>= 34650</p><p>When the four I occurs together we treat them as single object IIII. This single object together with 7 remaining object will account for 8 object which have 1-M. 2-P and 4-S.</p><p>So, required number of permutation =&nbsp;<span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>8</mn><mo>!</mo></mrow><mrow><mn>4</mn><mo>!</mo><mn>2</mn><mo>!</mo></mrow></mfrac></mrow></math></span><math><mrow><mfrac><mrow><mn></mn></mrow></mfrac></mrow></math></p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1747813013phpt8MTXP.jpeg" alt="" width="200" height="48" loading="lazy"></picture></div></div><p>= 840</p><p>Therefore, total no. of permutation in which 4-I's do not come together</p><p>= 34650 – 840</p><p>= 33810</p>

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